if (1 + x)^n = (n/0) + (n/1)x + (n/2)x^2 + ........... + (n/n)x^n
show by differentiation that
2(2/n) + (2x3)(n/3) + (3x4)(n/4) + ......... + n(n-1)(n/n) = n(n-1)2^n-2
You mean $\displaystyle (1+ x)^n= \left(\begin{array}{c}n \\ 0\end{array}\right)+ \left(\begin{array}{c}n \\ 1\end{array}\right)x+ \left(\begin{array}{c}n \\ 2\end{array}\right)x^2+ \cdot\cdot\cdot +\left(\begin{array}{c}n \\ n\end{array}\right)x^n$
(what you wrote looks too much like fractions!)
Have you tried this at all? The "$\displaystyle 2^{n-2}$" on the right makes me think of differentiating twice (so that the power is reduced to n-2) and then taking x= 1 (so that x+1= 1+ 1= 2). What do you get if you do that on both sides?
The equation we have is:
$\displaystyle (1+x)^n = \sum_{k=0}^n \binom{n}{k}x^k 1^{n-k} = \binom{n}{0}x^0 + \binom{n}{1}x + ... +\binom{n}{n-1}x^{n-1}+ \binom{n}{n}x^n$
Differentiating both sides once, we get:
$\displaystyle n(x+1)^{n-1} = \binom{n}{1} + 2\binom{n}{2}x + ... + (n-1)\binom{n}{n-1}x^{n-2} + n\binom{n}{n}x^{n-1}$
Differentiating again, we get:
$\displaystyle n(n-1)(x+1)^{n-2}= 2\binom{n}{2} + 2 \cdot 3\binom{n}{3}x + ... $ $\displaystyle + (n-1)(n-2)\binom{n}{n-1}x^{n-3} + n(n-1)\binom{n}{n}x^{n-2}$
If we take x = 1, we get:
$\displaystyle 2\binom{n}{2} + 2 \cdot 3\binom{n}{3} + ... + (n-1)(n-2)\binom{n}{n-1} + n(n-1)\binom{n}{n} = n(n-1)(1+1)^{n-2}$ $\displaystyle = n(n-1)2^{n-2}$