Find the sum `1.^(n)C_(0) + 3 .^(n)C_(1) + 5.^(n)C_(2) + "….." + (2n+1).^(n)C_(n)`.

Explanation

Method I :
`.^(n)C_(0)+2xx.^(n)C_(1)+3xx.^(n)C_(2)+"...."+(n+1)xx .^(n)C_(n)`
`= underset(r=0)overset(n)sum(r+1).^(n)C_(r)`
`=underset(r=0)overset(n)sum[r.^(n)C_(r)+.^(n)C_(r)]`
`=n underset(r=0)overset(n)sum.^(n-1)C_(r-1)+underset(r=0)overset(n)sum.^(n)C_(r)`
`= n(.^(n-1)C_(0) + .^(n-1)C_(1) + .^(n-1)C_(2)+"..."+.^(n-1)C_(n-1)) + (.^(n)C_(0)+.^(n)C_(1)+.^(n)C_(2)+"....."+.^(n)C_(n))`
`= n2^(n-1) + 2^(n)`
`= (n+2)2^(n-1)`
Method II :
We have `(1+x)^(n) = .^(n)C_(0)+.^(n)C_(1)x+.^(n)C_(2)x^(2)+"....."+.^(n)C_(n)x^(n)`
`:. x(1+x)^(n) = .^(n)C_(0)x+.^(n)C_(1)x^(2)+.^(n)C_(2)x^(3) + "....." + .^(n)C_(n)x^(n+1)`
Differentiating w.r.t. x, we get
`n(n1+x)^(n-1)x+(1+x)^(n)=.^(n)C_(0)+2xx.^(n)C_(1)x+
3xx.^(n)C_(2)x^(2)+"..."+(n+1)xx.^(n)C_(n)x^(n)`
Putting `x = 1`, we get
`n2^(n-1)+2^(n)=.^(n)C_(0) +2xx.^(n)C_(1)+3xx.^(n)C_(2)+"....."+(n+1)xx.^(n)C_(n)`