If `sum_(r=0)^(n)((r+2)/(r+1)).^n C_r=(2^8-1)/6` , then `n` is (A) 8 (B) 4 (C) 6 (D) 5

Options

A. 8
B. 4
C. 6
D. 5

Answer

D

Explanation

`underset(r=0)overset(n)sum((r+1)/(r+1)+(1)/(r+1)).^(n)C_(1)`
`= underset(r=0)overset(n)sum.^(n)C_(r)+underset(r=0)overset(n)sum(.^(n)C_(r))/(r+1)`
`= 2^(n)+(1)/(n+1).underset(r=0)overset(n)sum.^(n+1)C_(r+1)`
`= 2^(n)+(2^(n+1)-1)/(n+1)`
`= ((n+3)2^(n)-1)/(n+1)`
`= (2^(8)-1)/(6)`
`rArr n = 5`