If p is a real number and if the middle term in the expansion of (p2 +2)8 is 1120, find p .

This is a Long Answer Type Questions as classified in NCERT Exemplar Sol: Given expression is (P2+2)8 N u m b e r o f t e r m s = 8 + 1 = 9 ( o d d ) ∴ M i d d l e t e r m ( n + 1 2 ) t h t e r m = 9 + 1 2 = 1 0 2 = 5 t h t e r m T 5 = T 4 + 1 = C 4 8 ( P 2 ) 8 − 4 ( 2 ) 4 = C 4 8 P 4 2 4 × 2 4 = C 4 8 P 4 N o w C 4 8 P 4 = 1 1 2 0 ⇒ 8 × 7 × 6 × 5 4 × 3 × 2 × 1 . P 4 = 1 1 2 0 ⇒ 7 0 P 4 = 1 1 2 0 ⇒ P 4 = 1 1 2 0 7 0 = 1 6 ⇒ P 4 = 2 4 ⇒ P = ± 2 H e n c e , t h e r e q u i r e d v a l u e o f P = ± 2 .

If xp occurs in the expansion of (x2+1x)2n, prove that its coefficient is 4n-p32n+p3.

This is a Long Answer Type Questions as classified in NCERT Exemplar Sol: The given expression is (x2+1x)2n G e n e r a l T e r m T r + 1 = C r n x n − r y r = C r 2 n ( x 2 ) 2 n − r ( 1 x ) r = C r 2 n ( x ) 4 n − 2 r . 1 x r = C r 2 n ( x ) 4 n − 2 r − r = C r 2 n ( x ) 4 n − 3 r I f x p o c c u r s i n ( x 2 + 1 x ) 2 n t h e n 4 n − 3 r = p ⇒ 3 r = 4 n − p ⇒ r = 4 n − p 3 ∴ C o e f f i c i e n t o f x p = C r 2 n = C 4 n − p 3 2 n = ( 2 n ) ! ( 4 n − p 3 ) ! ( 2 n − 4 n − p 3 ) ! = ( 2 n ) ! ( 4 n − p 3 ) ! ( 6 n − 4 n + p 3 ) ! = ( 2 n ) ! ( 4 n − p 3 ) ! ( 2 n + p 3 ) ! H e n c e , c o e f f i c i e n t o f x p = ( 2 n ) ! ( 4 n − p 3 ) ! ( 2 n + p 3 ) !

Find the term independent of x , x ≠ 0 , in the expansion of (3x2/2 – 1/3x)15

This is a Short Answer Type Questions as classified in NCERT Exemplar Sol: G e n e r a l T e r m T r + 1 = C r n x n − r y r = C r 1 5 ( 3 x 2 2 ) 1 5 − r ( − 1 3 x ) r = C r 1 5 ( 3 2 ) 1 5 − r ( x ) 3 0 − 2 r ( − 1 3 ) r 1 x r = C r 1 5 ( 3 2 ) 1 5 − r ( x ) 3 0 − 2 r − r ( − 1 ) r 1 ( 3 ) r = C r 1 5 ( 3 2 ) 1 5 − r ( x ) 3 0 − 3 r ( − 1 ) r 1 ( 3 ) r f o r g e t t i n g t h e t e r m i n d e p e n d e n t o f x , 3 0 − 3 r = 0 ⇒ r = 1 0 On putting the value of r in the above expression, we get = C 1 0 1 5 ( 3 2 ) 1 5 − 1 0 ( − 1 ) 1 0 1 ( 3 ) 1 0 = C 1 0 1 5 ( 3 ) 5 ( 2 ) 5 . 1 ( 3 ) 1 0 = C 1 0 1 5 . 1 ( 2 ) 5 . ( 3 ) 5 = C 1 0 1 5 ( 1 6 ) 5 H e n c e , t h e r e q u i r e d t e r m = C 1 0 1 5 ( 1 6 ) 5

If the term free from x in the expansion of x – k/x2 10 is 405, find the value of k .

This is a Short Answer Type Questions as classified in NCERT Exemplar Sol: The given expression is (x−Kx2)10 G e n e r a l T e r m T r + 1 = C r n x n − r y r = C r 1 0 ( x ) 1 0 − r ( − K x 2 ) r = C r 1 0 ( x ) 1 0 − r 2 ( − K ) r ( 1 x 2 r ) = C r 1 0 ( x ) 1 0 − r 2 − 2 r ( − K ) r = C r 1 0 ( x ) 1 0 − r − 4 r 2 ( − K ) r = C r 1 0 ( x ) 1 0 − 5 r 2 ( − K ) r f o r g e t t i n g t h e t e r m i n d e p e n d e n t o f x , 1 0 − 5 r 2 = 0 ⇒ r = 2 On putting the value of r in the above expression, we get = C 2 1 0 ( − K ) 2 A c c o r d i n g t o t h e c o n d i t i o n o f t h e q u e s t i o n , w e h a v e C 2 1 0 K 2 = 4 0 5 ⇒ 1 0 . 9 2 . 1 K 2 = 4 0 5 ⇒ 4 5 K 2 = 4 0 5 ⇒ K 2 = 4 0 5 4 5 = 9 ∴ K = ± 3 H e n c e , t h e v a l u e o f K = ± 3

Find the middle term(s) in the expansion of: (x/a−ax)10. 3x – x3/6 9

This is a Short Answer Type Questions as classified in NCERT Exemplar Sol: (i)Given expression is (xa−ax)10 N u m b e r o f t e r m s = 1 0 + 1 = 1 1 ( o d d ) ∴ M i d d l e t e r m ( n + 1 2 ) t h t e r m = 1 1 + 1 2 = 1 2 2 = 6 t h t e r m G e n e r a l t e r m T r + 1 = C r n x n − r y r ⇒ T 5 + 1 = C 5 1 0 ( x a ) 1 0 − 5 ( − a x ) 5 = − C 5 1 0 x 5 a 5 . a 5 x 5 = − C 5 1 0 = − 1 0 × 9 × 8 × 7 × 6 5 × 4 × 3 × 2 × 1 . = − 9 × 7 × 4 = − 2 5 2 H e n c e , t h e r e q u i r e d m i d d l e t e r m = − 2 5 2 (ii)Given expression is (3x−x36)9 N u m b e r o f t e r m s = 9 + 1 = 1 0 ( e v e n ) ∴ M i d d l e t e r m s a r e ( n 2 ) t h t e r m a n d ( n 2 + 1 ) t h t e r m ( 1 0 2 ) t h = 5 t h t e r m a n d ( 1 0 2 + 1 ) t h = 6 t h t e r m G e n e r a l t e r m T r + 1 = C r n x n − r y r ∴ T 5 = T 4 + 1 = C 4 9 ( 3 x ) 9 − 4 ( − x 3 6 ) 4 = C 4 9 ( 3 ) 5 . x 5 ( − 1 6 ) 4 . x 1 2 = 9 × 8 × 7 × 6 4 × 3 × 2 × 1 × 3 × 3 × 3 × 3 × 3 6 × 6 × 6 × 6 x 1 7 = 1 8 9 8 x 1 7 N o w , T 6 = T 5 + 1 = C 5 9 ( 3 x ) 9 − 5 ( − x 3 6 ) 5 = C 5 9 ( 3 ) 4 . x 4 ( − 1 6 ) 5 . x 1 5 = 9 × 8 × 7 × 6 × 5 5 × 4 × 3 × 2 × 1 ( 3 ) 4 ( − 1 6 ) 5 . x 1 9 = − 2 1 1 6 x 1 9 H e n c e , t h e r e q u i r e d m i d d l e t e r m s a r e 1 8 9 8 x 1 7 a n d − 2 1 1 6 x 1 9

The sum of the series 10 20 r =0 Cr is 219 + 20 C10/2

This is a True or False Type Questions as classified in NCERT Exemplar Sol: ∑ r = 0 1 0 C r 2 0 = C 0 2 0 + C 1 2 0 + C 2 2 0 + C 3 2 0 + … + C 1 0 2 0 = C 0 2 0 + C 1 2 0 + … + C 1 0 2 0 + C 1 1 2 0 + … + C 2 0 2 0 − ( C 1 1 2 0 + … + C 2 0 2 0 ) = 2 2 0 − ( C 1 1 2 0 + … + C 2 0 2 0 ) H e n c e , t h e g i v e n s t a t e m e n t i s ' F a l s e '

In the expansion of x2 – 1/x2 16 , the value of the constant term is ________________ .

This is a Fill in the Blanks Type Questions as classified in NCERT Exemplar Sol: Let Tr+1 be the constant term in the expansion of (x2−1x2)16 ∴ T r + 1 = C r 1 6 ( x 2 ) 1 6 − r ( − 1 x 2 ) r = C r 1 6 ( x ) 3 2 − 2 r ( − 1 ) r . 1 x 2 r = ( − 1 ) r . C r 1 6 ( x ) 3 2 − 2 r − 2 r = ( − 1 ) r . C r 1 6 ( x ) 3 2 − 4 r F o r g e t t i n g c o n s t a n t t e r m , 3 2 − 4 r = 0 ⇒ r = 8 ∴ T r + 1 = ( − 1 ) 8 . C 8 1 6 = C 8 1 6 H e n c e , t h e v a l u e o f t h e f i l l e r i s C 8 1 6 .

If the seventh terms from the beginning and the end in the expansion of are equal, then n equals _________________ . [Hint: T7=Tn−7+2 , = nC6 (2 /3)n-6 (1/3 1/3 6) = nCn – 6 (2 1/3 6 (1/3 1/3) n - 6 solve for n. ] = (2 1/3)n – 12 = 1/3 1/3 n -12 = only problem when n – 12 = 0 = n =12].

This is a Fill in the Blanks Type Questions as classified in NCERT Exemplar Sol: The given expansion is (+1)n ∴ T 7 = T 6 + 1 = C 6 n ( 2 1 / 3 ) n − 6 . 1 ( 3 1 / 3 ) 6 = C 6 n ( 2 ) n − 6 3 ( − 1 ) r . 1 ( 3 ) 2 N o w t h e T 7 f r o m t h e e n d = T 7 f r o m t h e b e g i n n i n g i n ( 1 + ) n . ∴ T 7 = T 6 + 1 = C 6 n ( 1 3 1 / 3 ) n − 6 . ( 2 1 / 3 ) 6 A s p e r t h e q u e s t i o n s , w e g e t C 6 n ( 2 ) n − 6 3 . ( 1 3 2 ) = C 6 n . 1 3 n − 6 3 . ( 2 ) 2 ⇒ ( 2 ) n − 6 3 . ( 3 ) − 2 = ( 3 ) − ( n − 6 3 ) . ( 2 ) 2 ⇒ ( 2 ) n − 6 3 − 2 . ( 3 ) − 2 + n − 6 3 = 1 ⇒ ( 2 ) n − 1 2 3 . ( 3 ) n − 1 2 3 = 1 ⇒ ( 6 ) n − 1 2 3 = ( 6 ) 0 ⇒ n − 1 2 3 = 0 ⇒ n = 1 2 H e n c e , t h e v a l u e o f t h e f i l l e r i s 1 2 .

The coefficient of a−6b4 in the expansion of (1a+2b3)10 is _________________ . [Hint: T5 = 10C4 1/a b -2b/3 4 = 1120/27 a-6b4]

This is a Fill in the Blanks Type Questions as classified in NCERT Exemplar Sol: The given expansion is (1a−2b3)10 F r o m a − 6 b 4 , w e c a n t a k e r = 4 ∴ T 5 = T 4 + 1 = C 4 1 0 ( 1 a ) 1 0 − 4 . ( − 2 b 3 ) 4 = C 4 1 0 ( 1 a ) 6 . ( − 2 3 ) 4 . b 4 = 1 0 . 9 . 8 . 7 4 . 3 . 2 . 1 × 1 6 8 1 . a − 6 b 4 = 2 1 0 × 1 6 8 1 . a − 6 b 4 = 1 1 2 0 2 7 . a − 6 b 4 H e n c e , t h e v a l u e o f t h e f i l l e r i s 1 1 2 0 2 7 .

Maths NCERT Exemplar Solutions Class 11th Chapter Eight: Overview, Questions, Preparation

Q: The total number of terms in the expansion of (x+a)100+(x−a)100 after simplification is: (a) 50 (b) 202 (c) 51 (d) None of these

This is a Objective Type Questions as classified in NCERT Exemplar Sol: Number of terms in the expansion of (x+a)100=101 Number of terms in the expansion of (x−a)100=101 Now 50 terms of expansion will cancel out with negative 50 terms of (x−a)100 So, the remaining 51 terms of first expansion will be added to 51 terms of other. T h e r e f o r e , t h e n u m b e r o f t e r m s = 5 1 H e n c e , t h e c o r r e c t o p t i o n i s ( c ) .

Q: Given the integers r>1,n>2 , and the coefficients of the (3r) th and (r+2) nd terms in the binomial expansion of (1+x)2n are equal, then: (a) n=2r (b) n=3r (c) n=2r+1 (d) None of these

This is a Objective Type Questions as classified in NCERT Exemplar Sol: G i v e n t h a t r > 1 a n d n > 2 t h e n T 3 r = T 3 r − 1 + 1 = C 3 r − 1 2 n . x 3 r − 1 a n d T r + 2 = T r + 1 + 1 = C r + 1 2 n . x r + 1 A s p e r q u e s t i o n , w e h a v e C 3 r − 1 2 n = C r + 1 2 n ⇒ 3 r − 1 + r + 1 = 2 n [ ? C p n = C q n ⇒ n = p + q ] ⇒ 4 r = 2 n ⇒ n = 2 r H e n c e , t h e c o r r e c t o p t i o n i s ( a ) .

Q: The two successive terms in the expansion of (1+x)24 whose coefficients are in the ratio 1:4 are: (a) 3rd and 4th (b) 4th and 5th (c) 5th and 6th (d) 6th and 7th

This is a Objective Type Questions as classified in NCERT Exemplar Sol: Let rth and (r+1)th be two successive terms in the expansion (1+x)24 ∴ T r + 1 = C r 2 4 . x r a n d T r + 2 = T r + 1 + 1 = C r + 1 2 4 . x r + 1 A s p e r q u e s t i o n , w e h a v e C r 2 4 C r + 1 2 4 = 1 4 ⇒ 2 4 ! r ! ( 2 4 − r ) ! 2 4 ! ( r + 1 ) ! ( 2 4 − r − 1 ) ! = 1 4 ⇒ 2 4 ! r ! ( 2 4 − r ) ! × ( r + 1 ) ! ( 2 4 − r − 1 ) ! 2 4 ! = 1 4 ⇒ ( r + 1 ) . r ! ( 2 4 − r − 1 ) ! r ! ( 2 4 − r ) ( 2 4 − r − 1 ) ! = 1 4 ⇒ r + 1 2 4 − r = 1 4 ⇒ 4 r + 4 = 2 4 − r ⇒ 5 r = 2 0 ⇒ r = 4 ∴ T 4 + 1 = T 5 a n d T 4 + 2 = T 6 H e n c e , t h e c o r r e c t o p t i o n i s ( c ) .

Q: The coefficient of xn in the expansion of (1+x)2n and (1+x)2n−1 are in the ratio: (a) 1:2 (b) 1:3 (c) 3:1 (d) 2:1

This is a Objective Type Questions as classified in NCERT Exemplar Sol: G e n e r a l T e r m T r + 1 = C r n x n − r y r In the expansion of (1+x)2n, we get Tr+1=Cr2n.xr T o g e t t h e c o e f f i c i e n t s o f x n , p u t r = n ∴ C o e f f i c i e n t s o f x n = C n 2 n In the expansion of (1+x)2n−1, we get Tr+1=Cr2n−1.xr ∴ C o e f f i c i e n t s o f x n = C n − 1 2 n − 1 T h e r e q u i r e d r a t i o i s C n 2 n C n − 1 2 n − 1 = 2 n ! n ! ( n ) ! ( 2 n − 1 ) ! ( n − 1 ) ! ( 2 n − 1 − n + 1 ) ! = 2 n ! n ! n ! ( 2 n − 1 ) ! ( n − 1 ) ! ( n ) ! = 2 n ! n ! n ! × ( n − 1 ) ! n ! ( 2 n − 1 ) ! = 2 n ( 2 n − 1 ) ! n ! n ( n − 1 ) ! × ( n − 1 ) ! n ! ( 2 n − 1 ) ! = 2 1 = 2 : 1 H e n c e , t h e c o r r e c t o p t i o n i s ( d ) .

Q: If the coefficients of the 2nd, 3rd, and 4th terms in the expansion of (1+x)n are in A.P., then the value of n is: (a) 2 (b) 7 (c) 11 (d) 14 [Hint: 2 nC2 = nC1 + nC3 ⇒n2−9n+14=0⇒n=27. ]

This is a Objective Type Questions as classified in NCERT Exemplar Sol: Given expression is (1+x)n ( 1 + x ) n = C 0 n + C 1 n x + C 2 n x 2 + C 3 n x 3 + … + C n n x n H e r e , c o e f f i c i e n t o f 2 n d t e r m = C 1 n C o e f f i c i e n t o f 3 r d t e r m = C 2 n a n d c o e f f i c i e n t o f 4 t h t e r m = C 3 n G i v e n t h a t C 1 n , C 2 n a n d C 3 n a r e i n A . P ∴ 2 . C 2 n = C 1 n + C 3 n ⇒ 2 . n ( n − 1 ) 2 = n + n ( n − 1 ) ( n − 2 ) 3 . 2 . 1 ⇒ n ( n − 1 ) = n + n ( n − 1 ) ( n − 2 ) 6 ⇒ n − 1 = 1 + ( n − 1 ) ( n − 2 ) 6 ⇒ 6 n − 6 = 6 + n 2 − 3 n + 2 ⇒ n 2 − 3 n − 6 n + 1 4 = 0 ⇒ n 2 − 9 n + 1 4 = 0 ⇒ n 2 − 7 n − 2 n + 1 4 = 0 ⇒ n ( n − 7 ) − 2 ( n − 7 ) = 0 ⇒ ( n − 2 ) ( n − 7 ) = 0 ⇒ n = 2 , 7 ⇒ n = 7 w h e r e a s n = 2 i s n o t p o s s i b l e H e n c e , t h e c o r r e c t o p t i o n i s ( b ) .

Updated on Jul 23, 2025 15:30 IST

By Vishal Baghel, Executive Content Operations

CCCCC

Table of content

Binomial Theorem Long Answer Type Questions
Binomial Theorem Short Answer Type Questions
Binomial Theorem Objective Type Questions
Binomial Theorem Fill in the Blanks
Binomial Theorem True or False Type Questions
JEE Mains

Binomial Theorem Long Answer Type Questions

If $p$ is a real number and if the middle term in the expansion of ${(\frac{p}{2} + 2)}^{8}$ is 1120, find $p$ .

\begin{array}{l} G i v e n expression i s {(\frac{P}{2} + 2)}^{8} \\ N u m b e r o f t e r m s = 8 + 1 = 9 (o d d) \\ ∴ M i d d l e t e r m {(\frac{n + 1}{2})}^{t h} t e r m = \frac{9 + 1}{2} = \frac{10}{2} = 5^{t h} t e r m \\ T_{5} = T_{4 + 1} = {}_{8}{C_{4}} {(\frac{P}{2})}^{8 - 4} {(2)}^{4} \\ = {}_{8}{C_{4}} \frac{P^{4}}{2^{4}} \times 2^{4} = {}_{8}{C_{4}} P^{4} \\ N o w {}_{8}{C_{4}} P^{4} = 1120 \Rightarrow \frac{8 \times 7 \times 6 \times 5}{4 \times 3 \times 2 \times 1} . P^{4} = 1120 \\ \Rightarrow 70 P^{4} = 1120 \Rightarrow P^{4} = \frac{1120}{70} = 16 \\ \Rightarrow P^{4} = 2^{4} \Rightarrow P = \pm 2 \\ H e n c e, t h e r e q u i r e d v a l u e o f P = \pm 2 . \end{array}

Show that the middle term in the expansion of (x – 1/x)2n is 1 × 3 × 5 × … (2n -1)/n × (-2)n.

\begin{array}{l} G i v e n expression i s {(x - \frac{1}{x})}^{2 n} \\ N u m b e r o f t e r m s = 2 n + 1 = 9 (o d d) \\ ∴ M i d d l e t e r m {(\frac{n + 1}{2})}^{t h} t e r m = \frac{2 n + 1 + 1}{2} = (n + 1) t h t e r m . \\ G e n e r a l T e r m T_{r + 1} = {}_{n}{C_{r}} x^{n - r} y^{r} \\ ∴ T_{n + 1} = {}_{2 n}{C_{n}} {(x)}^{2 n - n} {(- \frac{1}{x})}^{n} = {}_{2 n}{C_{n}} {(x)}^{n} {(- 1)}^{n} . \frac{1}{x^{n}} \\ = {(- 1)}^{n} . {}_{2 n}{C_{n}} = {(- 1)}^{n} . \frac{2 n!}{n! (2 n - n)!} \\ = {(- 1)}^{n} . \frac{2 n!}{n! n!} = {(- 1)}^{n} . \frac{2 n (2 n - 1) (2 n - 2) (2 n - 3) \dots 1}{n! n (n - 1) (n - 2) (n - 3) \dots 1} \\ = {(- 1)}^{n} . \frac{2 n (2 n - 1) . 2 (n - 1) (2 n - 3) \dots 1}{n! n (n - 1) (n - 2) (n - 3) \dots 1} \\ = \frac{{(- 1)}^{n} . 2^{n} . [n (n - 1) (n - 1) \dots] . [(2 n - 1) . (2 n - 3) \dots 5.3.1]}{n! . n (n - 1) (n - 2) \dots 1} \\ = \frac{{(- 2)}^{n} [(2 n - 1) . (2 n - 3) \dots 5.3.1]}{n!} \\ = \frac{1 \times 3 \times 5 \times \dots (2 n - 1)}{n!} \times {(- 2)}^{n} \\ H e n c e, t h e m i d d l e t e r m = \frac{1 \times 3 \times 5 \times \dots (2 n - 1)}{n!} \times {(- 2)}^{n} \end{array}

Commonly asked questions

If $p$ is a real number and if the middle term in the expansion of ${(\frac{p}{2} + 2)}^{8}$ is 1120, find $p$ .

Show that the middle term in the expansion of (x – 1/x)²ⁿis 1 × 3 ×5 × … (2n -1)/n × (-2)ⁿ.

Find $n$ in the binomial expansion $,$ if the ratio of the 7th term from the beginning to the 7th term from the end is $\frac{1}{6} .$

In the expansion of ${(x + a)}^{n}$ if the sum of odd terms is denoted by $O$ and the sum of even terms by $E$ , prove that:

$O^{2} - E^{2} = {(x^{2} - a^{2})}^{n} .$

$4 O E = {(x + a)}^{2 n} - {(x - a)}^{2 n} .$

If $x^{p}$ occurs in the expansion of ${(\frac{x^{2} + 1}{x})}^{2 n},$ prove that its coefficient is $\frac{4 n - p}{3} \frac{2 n + p}{3} .$

Find the term independent of $x$ in the expansion of ${(1 + x + 2 x^{3})}^{}$ 3/2 x² – 1/3x ⁹.

Binomial Theorem Short Answer Type Questions

Find the term independent of x, $x \neq 0$ , in the expansion of (3x²/2 – 1/3x)¹⁵

\begin{array}{l} G e n e r a l T e r m T_{r + 1} = {}_{n}{C_{r}} x^{n - r} y^{r} \\ = {}_{15}{C_{r}} {(\frac{3 x^{2}}{2})}^{15 - r} {(- \frac{1}{3 x})}^{r} = {}_{15}{C_{r}} {(\frac{3}{2})}^{15 - r} {(x)}^{30 - 2 r} {(- \frac{1}{3})}^{r} \frac{1}{x^{r}} \\ = {}_{15}{C_{r}} {(\frac{3}{2})}^{15 - r} {(x)}^{30 - 2 r - r} {(- 1)}^{r} \frac{1}{{(3)}^{r}} \\ = {}_{15}{C_{r}} {(\frac{3}{2})}^{15 - r} {(x)}^{30 - 3 r} {(- 1)}^{r} \frac{1}{{(3)}^{r}} \\ f o r g e t t i n g t h e t e r m i n d e p e n d e n t o f x, \\ 30 - 3 r = 0 \Rightarrow r = 10 \\ O n p u t t i n g t h e v a l u e o f r i n t h e a b o v e expression, w e g e t \\ = {}_{15}{C_{10}} {(\frac{3}{2})}^{15 - 10} {(- 1)}^{10} \frac{1}{{(3)}^{10}} = {}_{15}{C_{10}} \frac{{(3)}^{5}}{{(2)}^{5}} . \frac{1}{{(3)}^{10}} \\ = {}_{15}{C_{10}} . \frac{1}{{(2)}^{5} . {(3)}^{5}} = {}_{15}{C_{10}} {(\frac{1}{6})}^{5} \\ H e n c e, t h e r e q u i r e d t e r m = {}_{15}{C_{10}} {(\frac{1}{6})}^{5} \end{array}

If the term free from in the expansion of x – k/x² ¹⁰is 405, find the value of .

\begin{array}{l} T h e g i v e n expression i s {(\sqrt[]{x} - \frac{K}{x^{2}})}^{10} \\ G e n e r a l T e r m T_{r + 1} = {}_{n}{C_{r}} x^{n - r} y^{r} \\ = {}_{10}{C_{r}} {(\sqrt[]{x})}^{10 - r} {(\frac{- K}{x^{2}})}^{r} = {}_{10}{C_{r}} {(x)}^{\frac{10 - r}{2}} {(- K)}^{r} (\frac{1}{x^{2 r}}) \\ = {}_{10}{C_{r}} {(x)}^{\frac{10 - r}{2} - 2 r} {(- K)}^{r} = {}_{10}{C_{r}} {(x)}^{\frac{10 - r - 4 r}{2}} {(- K)}^{r} \\ = {}_{10}{C_{r}} {(x)}^{\frac{10 - 5 r}{2}} {(- K)}^{r} \\ f o r g e t t i n g t h e t e r m i n d e p e n d e n t o f x, \\ \frac{10 - 5 r}{2} = 0 \Rightarrow r = 2 \\ O n p u t t i n g t h e v a l u e o f r i n t h e a b o v e expression, w e g e t \\ = {}_{10}{C_{2}} {(- K)}^{2} \\ A c c o r d i n g t o t h e c o n d i t i o n o f t h e q u e s t i o n, w e h a v e \\ {}_{10}{C_{2}} K^{2} = 405 \Rightarrow \frac{10.9}{2.1} K^{2} = 405 \\ \Rightarrow 45 K^{2} = 405 \Rightarrow K^{2} = \frac{405}{45} = 9 \\ ∴ K = \pm 3 \\ H e n c e, t h e v a l u e o f K = \pm 3 \end{array}

Commonly asked questions

Find the term independent of $x$ , $x \neq 0$ , in the expansion of (3x²/2 – 1/3x)¹⁵

If the term free from $x$ in the expansion of x – k/x² ¹⁰ is 405, find the value of $k$ .

Find the coefficient of $x$ in the expansion of

$(1 - 3 x + 7 x^{2}) {(1 - x)}^{16} .$

Find the term independent of $x$ in the expansion of 3x – 2/x^{2 15}

Find the middle term(s) in the expansion of:

${(x / a - \frac{a}{x})}^{10} .$ 3x – x³/6 ⁹

Find the coefficient of $x^{15}$ in the expansion of ${(x - x^{2})}^{10} .$

Find the coefficient of $\frac{1}{x^{17}}$ in the expansion of x⁴– 1/x^{3 15}

Find the sixth term of the expansion ${(y^{\frac{1}{2}} + x^{\frac{1}{3}})}^{n}$ , if the binomial coefficient of the third term from the end is 45.

[Hint: Binomial coefficient of third term from the end = Binomial coefficient of third term from the beginning = ⁿC₂.]

Find the value of $r$ , if the coefficients of the (2r + 4)^th and ^th terms in the expansion of ${(1 + x)}^{18}$ are equal.

If the coefficient of second, third, and fourth terms in the expansion of ${(1 + x)}^{2 n}$ are in A.P. Show that $2 n^{2} - 9 n + 7 = 0 .$

Find the coefficient of $x^{4}$ in the expansion of ${(1 + x + x^{2} + x^{3})}^{11} .$

Binomial Theorem Objective Type Questions

The total number of terms in the expansion of ${(x + a)}^{100} + {(x - a)}^{100}$ after simplification is:

(a) 50

(b) 202

(c) 51

(d) None of these

\begin{array}{l} N u m b e r o f t e r m s i n t h e expansion o f {(x + a)}^{100} = 101 \\ N u m b e r o f t e r m s i n t h e expansion o f {(x - a)}^{100} = 101 \\ N o w 50 t e r m s o f expansion w i l l c a n c e l o u t w i t h n e g a t i v e 50 t e r m s o f {(x - a)}^{100} \\ S o, t h e r e m a i n i n g 51 t e r m s o f f i r s t expansion w i l l b e a d d e d t o 51 t e r m s o f o t h e r . \\ T h e r e f o r e, t h e n u m b e r o f t e r m s = 51 \\ H e n c e, t h e c o r r e c t o p t i o n i s (c) . \end{array}

Given the integers $r > 1, n > 2$ , and the coefficients of the $(3 r)$ th and $(r + 2)$ nd terms in the binomial expansion of ${(1 + x)}^{2 n}$ are equal, then:

(a) $n = 2 r$

(b) $n = 3 r$

(c) $n = 2 r + 1$

(d) None of these

\begin{array}{l} G i v e n t h a t r > 1 a n d n > 2 \\ t h e n T_{3 r} = T_{3 r - 1 + 1} = {}_{2 n}{C_{3 r - 1}} . x^{3 r - 1} \\ a n d T_{r + 2} = T_{r + 1 + 1} = {}_{2 n}{C_{r + 1}} . x^{r + 1} \\ A s p e r q u e s t i o n, w e h a v e \\ {}_{2 n}{C_{3 r - 1}} = {}_{2 n}{C_{r + 1}} \\ \Rightarrow 3 r - 1 + r + 1 = 2 n [∵ {}_{n}{C_{p}} = {}_{n}{C_{q}} \Rightarrow n = p + q] \\ \Rightarrow 4 r = 2 n \\ \Rightarrow n = 2 r \\ H e n c e, t h e c o r r e c t o p t i o n i s (a) . \end{array}

Commonly asked questions

The total number of terms in the expansion of ${(x + a)}^{100} + {(x - a)}^{100}$ after simplification is:

(a) 50

(b) 202

(c) 51

(d) None of these

Given the integers $r > 1, n > 2$ , and the coefficients of the $(3 r)$ th and $(r + 2)$ nd terms in the binomial expansion of ${(1 + x)}^{2 n}$ are equal, then:

(a) $n = 2 r$

(b) $n = 3 r$

(c) $n = 2 r + 1$

(d) None of these

The two successive terms in the expansion of ${(1 + x)}^{24}$ whose coefficients are in the ratio 1:4 are:

(a) 3^rd and 4^th

(b) 4^th and 5^th

(c) 5^th and 6^th

(d) 6^th and 7^th

The coefficient of $x^{n}$ in the expansion of ${(1 + x)}^{2 n}$ and ${(1 + x)}^{2 n - 1}$ are in the ratio:

(a) 1:2

(b) 1:3

(c) 3:1

(d) 2:1

If the coefficients of the 2nd, 3rd, and 4th terms in the expansion of ${(1 + x)}^{n}$ are in A.P., then the value of $n$ is:

(a) 2

(b) 7

(c) 11

(d) 14

[Hint: 2 ⁿC₂ = ⁿC₁ + ⁿC₃ $\Rightarrow n^{2} - 9 n + 14 = 0 \Rightarrow n = 27 .$ ]

If $A$ and $B$ are coefficients of $x^{n}$ in the expansions of ${(1 + x)}^{2 n}$ and ${(1 + x)}^{2 n - 1}$ , respectively, then $\frac{A}{B}$

(a) 1

(b) 2

(c) $\frac{1}{2}$

(d) $\frac{1}{n}$

[Hint: $\frac{A}{B} =$ ²ⁿC_n/^{2n – 1}C_n = 2]

If the middle term of ${(1 + \frac{x sin x}{x})}^{10}$ is equal to $7 \frac{7}{8}$ , then the value of $x$ is:

(a) $2 n π + \frac{π}{6}$

(b) $n π + \frac{π}{6}$

(c) $n π + {(- 1)}^{n} \frac{π}{6}$

(d) $n π + {(- 1)}^{n} \frac{π}{3}$

[Hint: T₆ = ¹⁰C₅ 1/x₅ . x⁵ sin⁵ x = 63/3 = sin⁵ x = 1/25 sin ½= x = nπ + (-1)ⁿπ/6]

Binomial Theorem Fill in the Blanks

The largest coefficient in the expansion of ${(1 + x)}^{30}$ is _________________ .

\begin{array}{l} H e r e n = 30 w h i c h i s e v e n \\ ∴ t h e largest c o e f f i c i e n t i n {(1 + x)}^{n} = {}_{n}{C_{n / 2}} \\ S o, t h e largest c o e f f i c i e n t i n {(1 + x)}^{30} = {}_{30}{C_{15}} \\ H e n c e, t h e v a l u e o f t h e f i l l e r i s {}_{30}{C_{15}} . \end{array}

The number of terms in the expansion of ${(x + y + z)}^{n}$ is _________________

[Hint: ${(x + y + z)}^{n} = {[x + (y + z)]}^{n} .$ ]

\begin{array}{l} T h e expression {(x + y + z)}^{n} c a n b e w r i t t e n a {[x + (y + z)]}^{n} \\ ∴ {[x + (y + z)]}^{n} = {}_{n}{C_{0}} x^{n} {(y + z)}^{0} + {}_{n}{C_{1}} {(x)}^{n - 1} (y + z) + {}_{n}{C_{2}} {(x)}^{n - 2} {(y + z)}^{2} + \dots + {}_{n}{C_{n}} {(y + z)}^{n} \\ ∴ N u m b e r o f t e r m s 1 + 2 + 3 + 4 + \dots + (n + 1) = \frac{(n + 1) (n + 2)}{2} \\ H e n c e, t h e v a l u e o f t h e f i l l e r i s \frac{(n + 1) (n + 2)}{2} \end{array}

Commonly asked questions

The largest coefficient in the expansion of ${(1 + x)}^{30}$ is _________________ .

The number of terms in the expansion of ${(x + y + z)}^{n}$ is _________________

[Hint: ${(x + y + z)}^{n} = {[x + (y + z)]}^{n} .$ ]

In the expansion of x² – 1/x² ¹⁶ $,$ the value of the constant term is ________________ .

If the seventh terms from the beginning and the end in the expansion of are equal, then $n$ equals _________________ .

[Hint: $T_{7} = T_{n - 7 + 2}$ , = ⁿC⁶ (2 /3)^n-6 (1/3 ^{1/3 6}) = ⁿC^{n – 6} (2 ^1/3 ⁶ (1/3 ^1/3) ⁿ^{- 6} solve for $n .$ ]

= (2 ^1/3)ⁿ^{– 12} = 1/3 ^{1/3 n -12} = only problem when n – 12 = 0 = n =12].

The coefficient of $a^{- 6} b^{4}$ in the expansion of ${(\frac{1}{a} + \frac{2 b}{3})}^{10}$ is _________________ .

[Hint: T₅ = ¹⁰C₄ 1/a ^b -2b/3 ⁴ = 1120/27 a^-6b⁴]

The middle term in the expansion of ${(a^{3} + b a)}^{28}$ is _________________ .

The ratio of the coefficients of $x^{p}$ and $x^{q}$ in the expansion of ${(1 + x)}^{p + q}$ is _________________ .

[Hint: : ^p^{+ q}C_p = ^p^{+ q}C_q

The position of the term independent of $x$ in the expansion of $\sqrt {\frac{x}{3} + \frac{3}{{2 x}^{2}}}^{10}$ is _________________ .

If 2515 is divided by 13, the remainder is _________________ .

Binomial Theorem True or False Type Questions

The sum of the series ^{10 20} _{r =0} C_ris 2¹⁹ + ²⁰ C₁₀/2

\begin{array}{l} \sum_{r = 0}^{10} {}_{20}{C_{r}} = {}_{20}{C_{0}} + {}_{20}{C_{1}} + {}_{20}{C_{2}} + {}_{20}{C_{3}} + \dots + {}_{20}{C_{10}} \\ = {}_{20}{C_{0}} + {}_{20}{C_{1}} + \dots + {}_{20}{C_{10}} + {}_{20}{C_{11}} + \dots + {}_{20}{C_{20}} - ({}_{20}{C_{11}} + \dots + {}_{20}{C_{20}}) \\ = 2^{20} - ({}_{20}{C_{11}} + \dots + {}_{20}{C_{20}}) \\ H e n c e, t h e g i v e n s t a t e m e n t i s' F a l s e' \end{array}

The expression $7^{9} + 9^{7}$ is divisible by 64.

[Hint: $7^{9} + 9^{7} = {(1 + 8)}^{9} - {(1 - 8)}^{7}$ .]

\begin{array}{l} 7^{9} + 9^{7} = {(1 + 8)}^{7} - {(1 - 8)}^{9} \\ = [{}_{7}{C_{0}} + {}_{7}{C_{1}} . 8 + {}_{7}{C_{2}} {(8)}^{2} + {}_{7}{C_{3}} {(8)}^{3} + \dots + {}_{7}{C_{7}} {(8)}^{7}] - [\begin{array}{l} {}_{9}{C_{0}} - {}_{9}{C_{1}} . 8 + {}_{9}{C_{2}} {(8)}^{2} - {}_{9}{C_{3}} {(8)}^{3} \\ + \dots {}_{9}{C_{9}} {(8)}^{9} \end{array}] \\ = (7 \times 8 + 9 \times 8) + (21 \times 8^{2} - 36 \times 8^{2}) + \dots \\ = (56 + 72) + (21 - 36) \times 8^{2} + \dots \\ = 128 + 64 (21 - 36) + \dots \\ = 64 [2 + (21 - 36) + \dots] \\ w h i c h i s d i v i s i b l e b y 64 . \\ H e n c e, t h e g i v e n s t a t e m e n t i s' T r u e' . \end{array}

Commonly asked questions

The sum of the series ^{10 20} _{r =0} C_ris 2¹⁹ + ²⁰ C₁₀/2

The expression $7^{9} + 9^{7}$ is divisible by 64.

[Hint: $7^{9} + 9^{7} = {(1 + 8)}^{9} - {(1 - 8)}^{7}$ .]

The number of terms in the expansion of ${[{(2 x + y^{3})}^{4}]}^{7}$ is 8.

The sum of coefficients of the two middle terms in the expansion of ${(1 + x)}^{2 n - 1}$ is equal to ${}_{2 n - 1}{C_{n}}$ .

The last two digits of the number $3^{400}$ are 01.

If the expansion of ${(x - \frac{1}{x^{2}})}^{2 n}$ contains a term independent of $x$ , then $n$ is a multiple of 2.

The number of terms in the expansion of ${(a + b)}^{n}$ where $n \in ℕ$ is one less than the power $n$ .

JEE Mains

Maths NCERT Exemplar Solutions Class 11th Chapter Eight Exam

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Maths NCERT Exemplar Solutions Class 11th Chapter Eight: Overview, Questions, Preparation

Binomial Theorem Long Answer Type Questions

Commonly asked questions

Binomial Theorem Short Answer Type Questions

Commonly asked questions

Binomial Theorem Objective Type Questions

Commonly asked questions

Binomial Theorem Fill in the Blanks

Commonly asked questions

Binomial Theorem True or False Type Questions

Commonly asked questions

JEE Mains

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