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If `a_(1),a_(2),a_(3),…a_(n+1)` are in arithmetic progression, then `sum_(k=0)^(n) .^(n)C_(k.a_(k+1)` is equal to (a)`2^(n)(a_(1)+a_(n+1))` (b)`2^(n-1)(a_(1)+a_(n+1))` (c)`2^(n+1)(a_(1)+a_(n+1))` (d)`(a_(1)+a_(n+1))`

A

`2^(n)(a_(1)+a_(n+1))`

B

`2^(n-1)(a_(1)+a_(n+1))`

C

`2^(n+1)(a_(1)+a_(n+1))`

D

`(a_(1)+a_(n+1))`

Text Solution

Verified by Experts

The correct Answer is:
B

`sum_(k=0)^(n) .^(n)C_(k) a_(k-1)=sum_(k=0)^(n) .^(n)C_(k)(a_(1)+kd)`
`=a_(1)sum_(k=0)^(n) .^(n)C_(k)+dn=sum_(k=0)^(n) .^(n-1)C_(k-1)`
`=8_(1)2^(n)+nd.2^(n-1)`
`=2^(n-1)(2B_(1)+nd)`
`=2^(n-1)(a_(n+1)+a_(1))`
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