JEE Mains · Maths · STD 11 - 7. binomial theoram
Let \(\left(1+x+2 x^{2}\right)^{20}=a_{0}+a_{1} x+a_{2} x^{2}+\ldots+a_{40} x^{40}\) then \(a _{1}+ a _{3}+ a _{5}+\ldots+ a _{37}\) is equal to
- A \(2^{20}\left(2^{20}-21\right)\)
- B \(2^{19}\left(2^{20}-21\right)\)
- C \(2^{19}\left(2^{20}+21\right)\)
- D \(2^{20}\left(2^{20}+21\right)\)
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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