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JEE Mains · Maths · STD 11 - 7. binomial theoram

माना \(\left(1+x+2 x^{2}\right)^{20}=a_{0}+a_{1} x+a_{2} x^{2}+\ldots+a_{40} x^{40}\) है। तो \(a_{1}+a_{3}+a_{5}+\ldots+a_{37}\) बराबर है 

  1. A \(2^{20}\left(2^{20}-21\right)\)
  2. B \(2^{19}\left(2^{20}-21\right)\)
  3. C \(2^{19}\left(2^{20}+21\right)\)
  4. D \(2^{20}\left(2^{20}+21\right)\)
Verified Solution

Answer & Solution

Correct Answer

(B) \(2^{19}\left(2^{20}-21\right)\)

Step-by-step Solution

Detailed explanation

\(\left(1+x+2 x^{2}\right)^{20}=a_{0}+a_{1} x+\ldots .+a_{40} x^{40}\) put \(x=\) \(1,-1\) \(\Rightarrow a_{0}+a_{1}+a_{2}+\ldots .+a_{40}=2^{20}\) \(a_{0}-a_{1}+a_{2}+\ldots+a_{40}=2^{20}\) \(\Rightarrow a_{1}+a_{3}+\ldots+a_{39}=\frac{4^{20}-2^{20}}{2}\)…
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