(1+x)^20 + x(1+x)^19 + x^2 (1+x)^18 +....+ x^18 (1+x). prove that the coefficient of x^18 is 1330
I suspect a typo. The powers all add up to 20, except for the last term, which should probably be
x^18 (1+x)^2 + x^19(1+x)
or is there something else?
Okay. check a few terms
(x+1)^20 = x^20 + 20x^19 + (20*19)/2 x^18 + ...
x(x+1)^19 = x^20 + 19x^19 + (19*18)/2 x^18 + ...
x^2(x+1)^18 = x^20 + 18x^19 + (18*17)/2 x^18 + ...
So, the whole sum of the coefficients of x^18 is just
20
∑ k(k-1)/2
k=1
That is just
20
∑ 1/2 (k^2 - k)
k=1
You have formulas for sums of k and k^2, so just plug them in