Binomial Theorem

Let $n \geq 5$ be an integer. If 9ⁿ – 8n – 1 = 64a and 6ⁿ – 5n – 1 = 25b, then a - b is equal to

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Contributor-Level 10

$9^{n} - 8 n - 1 = {(1 + 8)}^{n} - 8 n - 1$

$= (1 + 8 n +^{n} ? C_{2} \cdot 8^{2} +^{n} ? C_{3} \cdot 8^{3} + . . . .] - 8 n - 1$

So, a = ⁿC₂ + ⁿC₃8 + ⁿC₄

8² +….

Similarly, b = ⁿC₂ + ⁿC₃ 5 + ⁿC₄ 5² +….

a - b = ⁿC₃ (8 – 5) + ⁿC₄ (8² – 5²) +….

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2 weeks ago

Let the coefficients of x^-¹ and x^-³ in the expansion of ${(2 x^{\frac{1}{5}} - \frac{1}{x^{\frac{1}{5}}})}^{15}, x > 0,$ be m and n respectively. If r is a positive integer such that mn² = ¹⁵C_r . 2^r, then the value of r is equal to….

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Contributor-Level 10

$T_{r + 1} =^{15} ? C_{r} \cdot {(2 x^{1 / 5})}^{15 - r} {(- x^{- 1 / 5})}^{r}$

= ¹⁵C_{r
$\cdot 2^{15 - r} \cdot x^{\frac{15 - 2 r}{5}} \cdot {(- 1)}^{r}$}

Coefficient of x^-1 -> r = 10 ->m = ¹⁵C_{10
$\cdot 2^{5}$}

$x^{- 3} \Rightarrow r = 15 \Rightarrow n =^{15} ? C_{15} \cdot 2^{0} = 1$

now mn² = ¹⁵C_{r
$\cdot 2^{r}$}

r = 5

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3 weeks ago

If the coefficient of x²¹ in the expansion of ${(x^{2} + \frac{3 λ}{x})}^{15}$ is 3640, then the value of $' λ'$ equal

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Contributor-Level 10

$T_{r + 1} =^{15} C_{r} {(x^{2})}^{15 - r} {(\frac{3 λ}{x})}^{r}$

$T_{r + 1} =^{15} C_{4} {(3 λ)}^{r} . x^{30 - 3 r}$

30 – 3r = 21 => r = 3

$λ^{3} = \frac{8}{27} \Rightarrow λ = \frac{2}{3}$

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3 weeks ago

If C₀, C₁, ……, C₂₀₁₂ are binomial coefficients in the expansion of (1 + x)²⁰¹² and a₀, a₁, …., a₂₀₁₂ are real numbers in arithmetic progression then value of

$a_{0} C_{0} - a_{1} C_{1} + a_{2} C_{2} - a_{3} C_{3} + . . . . . + a_{2012} C_{2012} i s a$

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Contributor-Level 10

$a_{0} C_{0} - a_{1} C_{1} + a_{2} C_{2} - a_{3} C_{3} + . . . . . + a_{2012} c_{2012}$

$= a_{0} (C_{0} - C_{1} + C_{2} + . . . . + C_{2012})$

$- d (C_{1} - 2 C_{2} + 3 C_{3} - 4 C_{4} + . . . . -^{2012} C_{2012})$

$= a_{0} (0) - d (C_{1} - 2 C_{2} + 3 C_{3} -^{4} C_{4})$

+ …….-

= $a_{0} (0) - d (0)$

= 0

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New answer posted

3 weeks ago

The remainder when 5²²²² is divided by 7 is:

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Contributor-Level 10

(5³)^740 * 5² = 5² (125)^740 = 5² (126 – 1)^740
So remainder = 4.

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3 weeks ago

The constant term in the expansion of (log₁₀(x^(log₁₀x)) - log_x² 100)¹² is:

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Contributor-Level 10

log x = t
(t² - 1/t)¹²
T? = ¹²C? (t²)¹²? (-1/t)? = ¹²C? t²? ³? (-1)?
For constant term r = 8 & coefficient ¹²C?
= (12*11*10*9) / 24 = 45*11 = 495

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3 weeks ago

The value of

$-^{15} C_{1} + 2 .^{15} C_{2} - 3 .^{15} C_{3} + . . . . - 15 .^{15} C_{15} +^{14} C_{1} +^{14} C_{3} +^{14} C_{5} + . . . . +^{14} C_{11} i s :$

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Contributor-Level 10

= A + B (say)

(1 + x)¹⁵ =

differentiating 15 (1 + x)¹⁴ = $\sum r .^{15} C_{r} x^{r - 1}$

put = x = -1

$\Rightarrow 0 = -^{15} C_{1} + 2 .^{15} C_{2} - . . . . . . . = A$

$B =^{14} C_{13} = 2^{13}$

$∴ A + B = 2^{13} - 14$

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New answer posted

3 weeks ago

Rakshit is allowed to select 1 or more books out of (2n + 1) distinct books. If the number of ways in which he may not select all of them is 126, then value of n is __________

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Contributor-Level 10

2²? ¹ - 1 - 1 = 126
2²? ¹ = 2?
2n + 1 = 7
n = 3

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3 weeks ago

Let S_n = ⁿC₀ + ⁿC₃ + ⁿC₆ + ⁿC₉ + ……, then the value of S_n, if n is not a multiple of 3, is

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Contributor-Level 10

(1 + x)? = C? + C? x + C? x² + . C? x?
Put x = 1
2? = C? + C? + C? + C? + … C?
Put x = ω
(1 + ω)? = C? + C? ω + C? ω² + . C? (ω)?
Put x = ω²
(1 + ω²)? = C? + C? ω² + C? ω² + … C? (ω²)?
∴ 1 + ω + ω² = 0
Now add all three equation (s)
2? + (1 + ω)? + (1 + ω²)? = 3 [C? + C? + C? + …]
C? + C? + C? + C? + …
= (2? + (1+ω)? + (1+ω²)? ) / 3
= (2? + (1+ω)? + (-ω)? ) / 3 = (2? - (-1)? ) / 3

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New answer posted

3 weeks ago

A possible value of 'x', for which the ninth term in the expansion of {3ˡᵒᵍ₃√⁽²⁵⁽ˣ⁻¹⁾⁺⁷⁾ + 3⁻¹/ˡᵒᵍ₃⁽⁵⁽ˣ⁻¹⁾⁺¹⁾}¹⁰ in the increasing powers of 3⁻¹/ˡᵒᵍ₃⁽⁵⁽ˣ⁻¹⁾⁺¹⁾ is equal to 180, is :

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