WBJEE · Maths · Binomial Theorem
If \(\left(1+x+x^2+x^3\right)^5=\sum_{k=0}^{15} a_k x^k\) then \(\sum_{k=0}^7(-1)^k \cdot a_{2 k}\) is equal to
- A \(2^5\)
- B \(4^5\)
- C 0
- D \(4^4\)
Answer & Solution
Correct Answer
(C) 0
Step-by-step Solution
Detailed explanation
\begin{aligned} & \text { Hint : } \sum_{k=0}^1(-1)^k a_{2 k} \\ & =a_0-a_2+a_4-a_6+a_8 \ldots . a_{14} \\ & \left(1+x+x^2+x^3\right)^5=a_0+a_1 x+\ldots . .+a_{15} x^{15} \\ & \text { put } x=i \\ & \left(1+i+i^2+i^3\right)^5=a_0+a_1 i+a_2 i^2+\ldots .+a_{15} i^{15}…
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