Prove that :`C_(0)-3C_(1)+5C_(2)- ………..(-1)^n(2n+1)C_(n)=0`

Explanation

The numberical value of last term of
` C_(n) - 3C_(1) + 5C_(2) - …+ (-1)^(n) (2n+1) C_(n) " is" (2n+1) C_(n)`
i.e. ` (2n+1)`
and `2n +1 = 2*n+1` or
` {:("n)(2n+ 1(2"),(" "underline(-2n)),(" "underline(1)):}`
Here , q = 2 and r = 1
The given series
` (1 + x)^(n) = C_(0) + C_(1)x + C_(2)x^(2) + C_(3) x^(3) + ...+ C_(n) x^(n)` now ,
replacing by `x^(2)` , then we get
`(1 + x^(2))^(n) = C_(0) + C_(1)x^(2) + C_(2)x^(4) + ...+ C_(n) x^(2n)`
On multiplying both sides by x , we get
` x (1 + x^(2))^(n) = C_(0)x + C_(1)x^(3) + C_(2) 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 = i in both sides , we get
` 0 + 0 = C_(0) - 3C_(1) + 5C_(2) - ...+ (2n+1) (-1)^(n) C_(n)`
or ` C_(0)- 3 C_(1) + 5C_(2) - ...+ (-1)^(n) (2n+1) C_(n) = 0 `
I Aliter
` LHS = C_(0) - 3C_(1) + 5C_(2) - ...+ (2n+1) (-1)^(n) C_(n)`
`= C_(0) - (1 + 2) C_(1) + (1 + 4) C_(2) - ...+ (-1)^(n) (1 + 2n)C_(n)`
`= (C_(0) - C_(1) + C_(2) - ...+ (-1)^(n)C_(n)) - 2n(C_(1) - 2C_(2) + ...+ (-1)^(n-1) n*C_(n))`
` = (1-1)^(n) - 2 *0 " " ` [ from Example ]
= 0 = RHS
II. Aliter
` LHS = C_(0) - 3C_(1) + 5C_(2) -...+ (-1)^(n) (2n+1) C_(n)`
`= sum_(r=1)^(n)(-1)^(r) (2r+1)""^(n)C_(r) = sum_(r=1)^(n) (-1)^(r) [2r*""^(n)C_(r) + ""^(n)C_(r)]`
` = 2 sum_(r=1)^(n) *""^(n-1)C_(r-1)+ sum_(r=1)^(n) (-1)^(r) *""^(n)C_(r)` .
` = 2n (1-n)^(n-1) + (1 - 1)^(n) = 0 + 0 = 0 = RHS `