The value of `C_(0)+3C_(1)+5C_(2)+7C_(3)+….+(2n+1)C_(n)` is equal to :

Options

A. `n.2^(n)`
B. `(n-1)2^(n)`
C. `(n+1)2^(n+1)`
D. `( n+1)2^(n)`

Answer

d

Explanation

We have,
`c_(0) + 3 C_(1) + 5 C_(2) + ...+(2n +1)C_(n)`
`sum_(r=1)^(n) (2r +1) C_(r)`
`sum_(r=1)^(n) (2r +1)""^(n)C_(r)" "[because C_(r) = ""^(n)C_(r)`]
`sum_(r=1)^(n) (2r. ""^(n)C_(r)+""^(n)C_(r))`
`sum_(r=1)^(n) 2r. ""^(n)C_(r)+sum_(r=1)^(n)""^(n)C_(r)`
`2sum_(r=1)^(n) r. ""^(n)C_(r)+sum_(r=1)^(n)""^(n)C_(r)`
`2sum_(r=1)^(n) r.(n)/(r) ""^(n-1)C_(r-1)+sum_(r=1)^(n)""^(n)C_(r)" "[because ""^(n)C_(r)= (n)/(r) .""^(n-1)C_(r-1)]`
`2nsum_(r=1)^(n) ""^(n-1)C_(r-1)+sum_(r=1)^(n)""^(n)C_(r)`
`2nsum_(r=1)^(n) 2^(n-1)+2^(n) = n.2^(n) + 2^(n) = (n-1)2^(n)` .