Class 11th

11. The general term of the expansion ${\left(1+x\right)}^m$ is given by,

T_r+1 = ^mC_r $1^{m-r} x^r$

= ^mC_rx^r

At r = 2,

T₂₊₁ = ^mC₂x²

Given that, co-efficient of x² = 6

=>^mC₂ = 6

=> $\frac{m!}{2!\left(m-2\right)!}$ = 6

=>m² – m = 12

=>m² – m – 12 = 0

=>m² + 3m – 4m – 12 = 0

=>m (m + 3) – 4 (m+ 3) = 0

=> (m – 4) (m + 3) = 0

=>m = 4 and m = –3

Since, we need a positive value of m we have,  m = 4

10. General term of the expansion (1 + x)²ⁿ is

T_r+1 = ²ⁿC_r (1)^2n-r(x)^r

So, co-efficient of xⁿ (i.e. r = n) is ²ⁿC_n

Similarly general term of the expansion (1 + x)^2n–1 is

T_r+1 = ^2n-1C_r (1)^2n–1–rx^r

And co-efficient of xⁿ i.e. when r = n is ^2n-1C_n

Therefore,

$\frac{co-efficient of x^n in {\left(1+x\right)}^{2n}}{co-efficient of x^n in {\left(1+x\right)}^{2n-1}}$

=

= $\frac{2n!}{n!\left(2n-n\right)!}$ ÷ $\frac{\left(2n-1\right)!}{n!\left(2n-1-n\right)!}$

= $\frac{2n!}{n!n!}$ * $\frac{n!\left(n-1\right)!}{\left(2n-1\right)!}$

= $\frac{2n}{n}$

= 2

Thus, co-efficient of $x^n$ in ${\left(1+x\right)}^{2n}$ = 2x co-efficient of $x^n$ in ${\left(1+x\right)}^{2n-1}$

8. The general term of the expansion (1 + a)^m+n is

T_r+1 = ^m+nC_ra^r   [since, 1^m+n-r = 1]

At r = m we have,

T_m+1 = ^m+nC_ma^m

= $\frac{\left(m+n\right)!}{m!\left(m+n-m\right)!}$ (a)^m

= $\frac{\left(m+n\right)!}{m! n!}$ a^m - (1)

Similarly at r = n we have,

T_n+1 = ^m+nC_naⁿ

= $\frac{\left(m+n\right)!}{n!\left(m+n-n\right)!}$ (a)ⁿ

= $\frac{\left(m+n\right)!}{n! m!}$ aⁿ - (2)

Hence from (1) & (2),

Co-efficient of am = Co-efficient of aⁿ = $\frac{\left(m+n\right)!}{m!n!}$

6. Let (r + 1)th be the general term of ( $x^2- yx){}^{12}$

So, T_r-1 = ¹²C_r (x²)^12–r (–yx)^r

= (–1)^r12C_rx^24–2ry^rx^r

= (–1)^r12C_r $x^{24-2r+r}{y}^r$

= (-1)^r12C­_r $x^{24-r}{y}^r$

5. Let (r + 1)^th term be the general term of (x²–y)⁶.

So, T_r-1 = ⁶C_r (x²)^6-r (-y)^r

= (–1)^r .⁶C_r . $x^{12-r}$ . $y^r$

4. Let a⁵b⁷ occurs in (r + 1)^th term of [removed]a – 2b)¹².

Now, T_r+1 = ¹²C_ra^12-r (–2b)^r

= (–1)^r12C_ra^12–r . 2^r. b^r

Comparing indices of a and b in T_r-1 with a⁵ and b⁷ we get,  r = 7

So, co-efficient of a⁵b⁷ is (–1)⁷¹²C₇ 2⁷