Coefficeint of x 40 in the expansion 1- x 240 x 2-2-1- x 2-5 is -A- 1B- 30 C 15C- 30 C 20D- None

The correct option is B 30C15 (1+x^2)^40( x^2 + 2 + 1/x^2)^ 5 = (1 + x^2)^40( (x^2 + 1)^2/x^2)^ 5 = (1 + x^2)^40x^10/(x^2 + 1)^10 = x^10 ( 1 + x^2)^30 ...

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Byju's Answer

Standard XII

Mathematics

General Term of Binomial Expansion

Coefficeint o...

Question

Coefficeint of $x^{40}$ in the expansion $(1 + x^{2})^{40} {(x^{2} + 2 + \frac{1}{x^{2}})}^{- 5}$ is :

³⁰C₁₅

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³⁰C₂₀

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None

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Solution

The correct option is A
³⁰C₁₅

$(1 + x^{2})^{40} {(x^{2} + 2 + \frac{1}{x^{2}})}^{- 5} = (1 + x^{2})^{40} {(\frac{(x^{2} + 1)^{2}}{x^{2}})}^{- 5} = (1 + x^{2})^{40} \frac{x^{10}}{(x^{2} + 1)^{10}}$

= $x^{10} (1 + x^{2})^{30}$

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Coefficeint of x40 in the expansion (1+x2)40(x2+2+1x2)−5 is :

The correct option is A 30C15 (1+x2)40(x2+2+1x2)−5=(1+x2)40((x2+1)2x2)−5=(1+x2)40x10(x2+1)10 = x10(1+x2)30

Coefficeint of $x^{40}$ in the expansion $(1 + x^{2})^{40} {(x^{2} + 2 + \frac{1}{x^{2}})}^{- 5}$ is :

The correct option is A
³⁰C₁₅

$(1 + x^{2})^{40} {(x^{2} + 2 + \frac{1}{x^{2}})}^{- 5} = (1 + x^{2})^{40} {(\frac{(x^{2} + 1)^{2}}{x^{2}})}^{- 5} = (1 + x^{2})^{40} \frac{x^{10}}{(x^{2} + 1)^{10}}$

= $x^{10} (1 + x^{2})^{30}$