Let a0,a1,a2……an∈R be in an arithmetic progression and let C0,C1,C2……Cn be binomial coefficients. Then ∑k=0nak.Ck=

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Let $a_{0}, a_{1}, a_{2} \dots \dots a_{n} \in R$ be in an arithmetic progression and let
$C_{0}, C_{1}, C_{2} \dots \dots C_{n}$ be binomial coefficients. Then $\sum_{k = 0}^{n} a_{k} . C_{k} =$

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$(a_{0} + a_{n}) . 2^{n - 1}$

$\frac{1}{2} (a_{0} + a_{n})$

$(a_{0} + a_{n})$

answer is B.

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Detailed Solution

$\begin{array}{rcl} S & = & a_{0} C_{0} + a_{1} C_{1} + a_{2} C_{2} + . . . . . + a_{n} C_{n} \\ S & = & a_{0} C_{n} + a_{1} C_{n - 1} + a_{2} C_{n - 2} + . . . . . + a_{n} C_{0} \\ 2 S & = & (a_{0} + a_{n}) C_{0} + (a_{1} + a_{n - 1}) C_{1} + (a_{2} + a_{n - 2}) C_{2} \\ Since a_{0}, a_{1}, a_{2} . . . . . . a_{n} are in A . P . \\ ∴ a_{0} + a_{n} & = & a_{1} + a_{n - 1} = a_{n} + a_{n - 2} = . . . . . . \\ 2 S & = & (a_{0} + a_{n}) [C_{0} + C_{1} + C_{2} + . . . . . . C_{n}] \\ 2 S & = & [a_{0} + a_{n}] 2^{n} \\ S & = & [a_{0} + a_{n}] 2^{n - 1} \end{array}$