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

Options

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))`

Answer

B

Explanation

`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))`