If `f(x)=sum_(r=1)^(n) { r^(2) (""^(n)C_(r)- ^(n) C_(r-1))+ (2r+1) ^(n) C_(r)}`
and `f(30)=30(2)^(lambda),` then the value of `lambda` is

Options

A. 3
B. 4
C. 5
D. 6

Explanation

Here, `f(x) = sum_(r=1)^(n){r^(2)(""^(n)C_(r)-^(n) C_(r-1)) + (2r+1)^(n)C_(r)}`
`= sum _(r=1)^(n) (r^(2)+2r+1)""^(n)C_(r)-r^(2)cdot ^(n) C_(r-1)`
`= sum _(r=1)^(n) ((r+1)^(2)cdot""^(n)C_(r)-r^(2)cdot ^(n) C_(r-1))`
`=(n+1)^(2) cdot ^(n)C_(n)-1^(2)cdot ^(n)C_(0)`
`= (n+1)^(2) -1 = (n^(2) + 2n)`
`therefore f(30)=(30)^(2) + 2(30)=960`
`=30xx 32 = 30 (2)^(5)=30(2)^(lambda)` [given]
Hence, `lambda = 5`