If `(1+x)^n=c_0+c_1x+c_2x^2+...+c_nx^n` then the value of `c_0+3c_1+5c_2+....+(2n+1)c_n` is-

Explanation

Here , last term of ` C_(0) + 3C_(1) + 5C_(2) + …+ (2n+1) C_(n)` is
`(2n+1) C_(n) "i.e., " (2n +1)` and last term with positive sign
Then , ` 2n + 1 = n* 2 +1`
or `{:("n)n+1(1"),(" "underline(-n)),(" "underline(1)):}`
Here , q = 2 and r = 1
The given series is
` (1 + x)^(n) = C_(0) + C_(1)x + C_(2) x^(2) + ...+ C_(n) x^(n)`
Now , replacing x by `x^(2)` , we get
` (1 + x^(2))^(n) = C_(0) + C_(1) x^(2) + C_(2) + C_(2)x^(4)+ ...+ C_(n)x^(2n)`
On multiplying bith sides by ` x^(1)` , we get
`x(1+x^(2))^(n) = C_(0) x + C_(1) x^(3) + C_(2)x^(5) + ...+ C_(n) x^(2n +1)`
On differentiating both sides w.r.t..x, we get
` x* n(1 + x^(2))^(n-1)*2x+ (1 + x^(2))^(n)*1 = C_(0) + 3C_(1) x^(2) + 5C_(2) x^(4) + ... + (2n +1) C_(n) x^(2n)`
Putting x = 1 , we get
`n*2^(n-1)* 2 + 2^(n) = C_(0) + 3C_(1) + 5C_(2) + ...+ (2n +1)C_(n)`
or ` C_(0) + 3C_(1) + 5C_(2) + ..+ (2n+1)C_(n) = (n+1) 2^(n)`
I .Aliter
LHS ` = C_(0) + 3C_(1) + 5C_(2) + ...+ (2n+1)C_(n)`
` = C_(0) + (1+2) C_(1) + (1 + 2) C_(2) + ..+ (1 + n)C_(n)` [use example]
` = 2^(n) + n.2^(n-1) = (n +2) 2^(n-1) ` RHS
II . Aliter
LHS `= C_(0) + 2C_(1) + 3C_(2) + ...+ (n+1) C_(n)`
`sum_(r=1)^(n+1) r. ""^(n)C_(r-1) = sum_(r=1)^(n+1) (r - 1 + 1) . ""^(n)C_(r-1)`
` = sum_(r=1)^(n+1) (r-1) . ""^(n)C_(r-1)+ ""^(n)C_(r-1)`
` [ because ""^(n)C_(r-1) = (n)/(r-1) . ""^(n-1)C_(r-2)]`
` = n(0 + ""^(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))`
`= n.2^(n-1) + 2^(n) = (n+ 2). 2^(n-1) = RHS `