`(n+2)nC_0(2^(n+1))-(n+1)nC_1(2^(n))+(n)nC_2(2^(n-1))-....` is equal to

Options

A. 4
B. 4n
C. 4(n+1)
D. 2(n+2)

Answer

C

Explanation

`t_(r+1)=(-1)^(r)(n-r+2).^(n)C_(r)2^(n-r+1)`
`= (n+2)2^(n+1)(-1)^(r).^(n)C_(r)(1/2)^(r)-2^(n+1)r.^(n)C_(r)(1/2)^(r)`
`= (n+2)2^(n+1).^(n)C_(r)(-1/2)^(r) + 2^(n)n.^(n-1)C_(r-1)(-1/2)^(r-1)`
`:.` Sum `= (n+2)2^(n+1){.^(n)C_(0) - .^(n)C_(1) xx 1/2 .^(n)C_(2)xx(1/2)^(2)-"...."}`
`+2^(n){.^(n-1)C_(0)-.^(n-1)C_(1)xx1/2+.^(n-1)C_(2)xx(1/2)^(2)+"...."}`
`= (n+2)2^(n+1)(1-1/2)^(n)+n2^(n)(1-1/2)^(n-1)`
`= 2(n+2)+2n`
`= 4n+4`