If `sum_(r=0)^(10)((r+2)/(r+1))^^n C_r=(2^8-1)/6` , then `n` is `8` b. `4` c. `6` d. `5`

Answer

5

Explanation

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