AP EAMCET · Maths · Binomial Theorem
Let \(a_0, a_1, a_2, \ldots a_n \in R\) be in an arithmetic progression and let \(C_0, C_1, C_2, \ldots, C_n\) be the binomial coefficients. Then \(\sum_{k=0}^n a_k \cdot C_k=\)
- A \(\frac{1}{2}\left(a_0+a_n\right)\)
- B \(\left(a_0+a_n\right) \cdot 2^{n-1}\)
- C \(\left(a_0+a_n\right)\)
- D 0
Answer & Solution
Correct Answer
(B) \(\left(a_0+a_n\right) \cdot 2^{n-1}\)
Step-by-step Solution
Detailed explanation
\[ \begin{aligned} \sum_{k=0}^n a_k \cdot C_k=a_0 C_0+a_1 C_1+a_2 C_2 & +\ldots+a_n C_n \\ =a_0 C_0+\left(a_0+d\right) C_1 & +\left(a_0+2 d\right) C_2 \\ & +\ldots+\left(a_0+n d\right) C_n \end{aligned} \] Where \(d\) is an common difference of an AP…
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