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Binomial theorem for positive integral Index

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Question

Suppose $a_{0} = 2017, a_{1} a_{2}, \dots, a_{n - 1}, 2023$ $= a_{n}$ an are in $A . P$ . Let $S = \frac{1}{2^{n + 1}} \sum_{r = 0}^{n} (^{n} C_{r}) a_{r} - 1000$
Then $S$ is equal to

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Solution

Put $^{n} C_{r} = C_{r}$ and let
$T = a_{0} C_{0} + a_{1} C_{1} + a_{2} C_{2} + \dots + a_{n} C_{n}$
Using, $C_{r} = C_{n - r} C_{r} = C_{n - r}$
$T = a_{n} C_{0} + a_{n - 1} C_{1} + \dots + a_{0} C_{n}$
Adding $(1)$ and $(2)$ , we get
$2 T = (a_{0} + a_{n}) C_{0} + (a_{1} + a_{n - 1}) C_{1} + (a_{2} + a_{n - 2}) C_{2} + \dots + (a_{n} + a_{0}) C_{n}$
But $a_{0} + a_{n} = a_{1} + a_{n - 1} = a_{2} + a_{n - 2} = \dots = a_{n} + a_{0} = 2017 + 2023$
$\begin{array}{r} ∴ 2 T = 4040 (C_{0} + C_{1} + \dots + C_{n}) \\ \Rightarrow T = (2020) 2^{n} = (1010) 2^{n + 1} \end{array}$
Thus, $S = \frac{1}{2^{n + 1}} (T) - 1000 = 10$

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