`(1 + x)^(n) = C_(0) + C_(1) x + C_(2) x^(2) + C_(3) x^(3) + … + C_(n) x^(n)` , prove
that ` C_(0) - 2C_(1) + 3C_(2) - 4C_(3) + … + (-1)^(n) (n+1) C_(n) = 0 `

Explanation

Numerical value of last term of
` C_(0) - 2C_(1) + 3C_(2) - 4C_(3) + …(-1)^(n) (n+1) C_(n)` is
`(n+1) C_(n) "I.e., " (n +1) ` , then
`n+ 1 = n*1 + 1 `
`{:("n)n(1"),(" "underline(-n)),(" "underline0):}`
Here , q = 1 and r = 1
The given series is
` (1 + x)^(n) = C_(0) + C_(1) x + C_(2) x^(2) + C_(3) + ...+ C_(n) x^(3)`
On multiplying both sides by x , we get
` x (1 + x)^(n) = C_(0) x + C_(1) x + C_(1) x^(2) + C_(2) x^(3) + C_(3) x^(4) + ...x C_(n) x^(n+1)`
On differentiating both sides w.r.t.x, we get
` x *n (1 + x)^(n-1) + (1 + x)^(n) * 1 = C_(0) + 2C_(1) x + 3C_(2) x^(2)`
` = 4C_(3) x^(3) + ... + (n+1) C_(n) x^(n)`
Putting x = - 1 , we get
` 0 = C_(0) - 2C_(1) + 3C_(2) - 4C_(3) + ... + (-1)^(n)(n+1) C_(n)`
or ` C_(0) - 2C_(1) + 3C_(2) + 4C_(3) + ... + (-1)^(n) (n+1) C_(n)`
I Aliter
`LHS = C_(0) - 2C_(1) + 3C_(2) - 4C_(3) + ...+ (-1)^(n) (n+1) C_(n)`
` = C_(0) - (C_(1) + C_(1)) + (C_(2) + 2C_(2)) - (C_(3) + 3C_(3)) + ...+ (-1)^(n) { C_(n) + n C_(n)} `
` = {C_(0) - C_(1) + C_(2) - C_(3) + ...+ (-1)^(n) C_(n)}`
` + {(- C_(1) + 2C_(2) - 3C_(2) - 3C_(3) + ...+ (-1)^(n) n C_(n)}`
`= (1-1)^(n)+ {{:(-n+ 2*(n(n-1))/(1*2) - 3 (n(n-1)(n-2))/(1*2*3)),(+ ...+ (-1)^(n) *n):}}`
`= n{""^(N)C_(0)- ""^(N)C_(1)+""^(N)C_(2)- ...+ (-1)^(N)""^(N)C_(N)}`
` = n(1-1)^(N) = 0 = RHS `
II.Aliter
LHS `= C_(1) - 2C_(2) + 3C_(3)- ...+ (-1)^(n-1) . nC_(n)`
` = sum_(r=1)^(n) (-1)^(r-1). r . ""^(r)C_(r)`
`= sum_(r=1)^(n) (-1)^(r-1)*n*""^(n-1)C_(r-1) [ because ""^(n)C_(r)= (n)/(r) *""^(n-1)C_(r-1)]`
`= nsum_(r=1)^(n) (-1)^(r-1)*^(n-1)C_(r-1)`
` = n(1-1)^(n-1)*= 0= RHS `