If `(1+x)^n=c_0+C_1x+C_2x^2++C_n x^n , using derivtives prove t h a t` `C_1-2C_2+3C_3++(-1)^(n-1)nC_n=0`

Explanation

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