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

Explanation

Here , last term of
` 1^(2) *C_(2)+ (2*3)C_(3) + …+ {(n-1)*n} C_(n) "is" (n-1) nC_(n)`
i.e. ` (n-1)n`
[start with greatest fector here greater fector is n] and last
term with positive sing , then ` n = n *1 + 0`
or `{:("n)n(1"),(" "underline(-n)),(" "underline0):}`
Here , q =1 and r = 0
The given series is
` (1 + x)^(n) = C_(0) + C_(1) x + C_(2) + C_(3) x^(3) + ....+ n C_(n) x^(n-1)`
Differentiating on both sides w.r.t.x, we get
Again , differentiating on both sides w.r.t.x, we get
` n(n-1) (1 + x)^(n-2) = 0 + 0 (1*2)C_(2) + (2*3) C_(3) x + ...+ {(n-1)*n } C_(n) x^(n-2)`
Putting x = 1 , we get
` n(n-1) (1 +1)^(n-2) = (1*2) C_(2) + (2*3) C_(3) + ...+ {(n-1)n} * C_(n)`
or ` (1*2) C_(2) + (2*3) C_(3) + ...+ {(n-1)n*}* C_(n) = n (n-1) 2^(n-2)`
I .Aliter
` LHS = (1*2) C_(2) + (2*3) C_(3) + (3*4) C_(4) + ...+ {(n-1)n} *C_(n)`
` = (1*2) (n(n-1))/(1*2) + (2*3) (n(n-1)(n-2))/(1*2*3) `
`+ (3*4) (n(n-1)(n-2)(n-3))/(1*2*3*4) + ...+ (n-1)n*1`
` = n(n-1) {1 + ((n-2))/(1) + ((n-2)(n-3))/(1*2) + .. + 1}`
Now , in bracket , let n - 2 = N , then
` = n(n-1) {1+ (N)/(1) + (N (N-1))/(2!) + ...+ }`
`= n (n-1){""^(N)C_(0) + ""^(N)C_(1) + ...+ ""^(N)C_(N)}`
` = n(n-1 )2^(n) = n(n-1) 2^(n-2) = RHS `
II. Aliter
` LHS = (1*2) C_(2) + (2*3) C_(3) + ...+ {(n-1)}C_(n)`
` = sum_(r=2)^(n) (r-1) *r * ""^(n)C_(r)`
` = sum_(r=2)^(n) (r-1) *r* (n)/(r) * (n-1)/((r-1)) * ""^(n-2)C_(r-2)`
` = (n-1)n sum _(r=2)^(n) ""^(n-2)C_(r-2)` .
` = (n-1)n(""^(n-2)C_(0) + ""^(n-2)C_(0) + ""^(n-2)C_(2) + ...+ ""^(n-2)C_(n-2))`
` = (n-1) n* 2^(n-2) = RHS `