If `(1+ x)^(n) = C_(0) + C_(1) x + C_(2)x^(2) + ...+ C_(n)x^(n)` , prove that
` C_(1) + 2C_(2) + 3C_(3) + ...+ n""C_(n) = n*2^(n-1)`

Explanation

Here , last term of `C_(1) + 2C_(2) + 3C_(3) + …+ nC_(n) "is " nC_(n) ` i.e., n
and n ` n= n *1 + 0 "or" n)n(1`
`{:(underline(n)),(underline(0)):}`
Here , q = 1 and r = 0
Then , the given series is
`(1 + x)^(n) = C_(0) + C_(1)x+C_(2) x^(2) + C_(3) x^(3) + ...+ C_(n) x^(n)`
Diiferentiating both sides w.r.t.x,we get
`n(1 + x)^(n-1) = 0+C_(1) + 2C_(2) + 3C_(3) x^(3) + ...+ nC_(n) x^(n-1)`
Putting x= 1 , we get
` n*2^(n-1) = C_(1) + 2C_(2) + 3C_(3) +...+ nC_(n)`
I . Aliter
` C_(1) + 2C_(2) + 3C_(3) + ...+ nC_(n)`
`= n+ 2.(n(n-1))/(1*2) + 3* (n(n-1)(n-2))/(1*2*3) + ...+ n*1`
`= n{1 + (n-1) + ((n-1)(n-2))/(1*2) + ...+ 1}`
Let n- 1 = N , then
LHS `= (1 + N) { 1+ N + (N(N -1))/(1*2) + ... + 1}`
` = (1 + N){1 +""^(N)C_(1) + ""^(N)C_(2) + ...+ ""^(N)C_(N) `
` = (1 + N)2^(N) = n*2^(2-1)` = RHS
II . Aliter
LHS = `C_(1) + 2C_(2) + 3 C_(3)+ ...+ nC_(n) = sum_(r=1)^(n) r* ""^(n)C_(r)`
`= n sum_(r=1)^(n)r.(n)/(r). ""^(n-1)C_(r-1)" "[ because ""^(n)C_(r)= (n)/(r). ""^(n-1)C_(r-1)]`
` n sum_(r=1)^(n) ""^(n-1)C_(r-1)`
` = n (""^(n-1)C_(0) + ""^(n+1)C_(1) + ""^(n-1)C_(2) + ...+ ""^(n-1)C_(n-2) + ...+ ""^(n-1)C_(n-1))`
`n*n^(n-1)` = RHS