Suppose a0=2017,a1a2,…,an−1,2023=an an are in A.P. Let S=12n+1∑r=0n  nCrar−1000Then S is equal to

# Suppose ${a}_{0}=2017,{a}_{1}{a}_{2},\dots ,{a}_{n-1},2023$$={a}_{n}$ an are in $A.P$. Let Then $S$ is equal to

1. A

10

2. B

40

3. C

2013

4. D

2021

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### Solution:

Put  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  and $\left(2\right)$, we get

$2T=\left({a}_{0}+{a}_{n}\right){C}_{0}+\left({a}_{1}+{a}_{n-1}\right){C}_{1}+\left({a}_{2}+{a}_{n-2}\right){C}_{2}+\dots +\left({a}_{n}+{a}_{0}\right){C}_{n}$

But ${a}_{0}+{a}_{n}={a}_{1}+{a}_{n-1}={a}_{2}+{a}_{n-2}=\dots ={a}_{n}+{a}_{0}=2017+2023$

$\begin{array}{r}\therefore 2T=4040\left({C}_{0}+{C}_{1}+\dots +{C}_{n}\right)\\ ⇒T=\left(2020\right){2}^{n}=\left(1010\right){2}^{n+1}\end{array}$

Thus, $S=\frac{1}{{2}^{n+1}}\left(T\right)-1000=10$

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