AIME 1986 Problem - Polynomials

Question

OBSERVATIONS

Let $$f(x)=1-x+x^2...+x^{16}-x^{17}$$ If we try to replace $x=y-1$ then we see that $y^2$ appears in every term after $1-x$ in $f(x)$.

By Binomial Expansion/observation Coefficient of $y^2$ is Σn-1 in $x^n$ simplified.

This sums our value of required coefficients to Σ of Σn-1 for n=1 to 17. The problem I am facing is to simplify this double Σ.

Given Answer: 816

Answer 1

In another answer and the comments, a simpler approach has been pointed out. I want to show how to continue with your (correct and intuitive) approach.

Following your idea, we have that $$f(y)=\sum_{n=0}^{17}(-1)^n\sum_{k=0}^n\binom nky^k(-1)^{n-k}$$ We are only interested in the summands with $k=2$, so we get $$a_2=\sum_{n=2}^{17}(-1)^n\binom n2(-1)^{n-2}=\sum_{n=2}^{17}\binom n2$$ Now you can write $\binom n2=\frac12n^2-\frac12n$ and use formulas for the first $n$ natural numbers and the first $n$ squares to arrive at the result.

Answer 2

Note that $$f(x)=1-x+x^2...+x^{16}-x^{17}=\frac{1-x^{18}}{x+1}.$$ So, $$f(y-1)=\frac{1-(y-1)^{18}}{y}.$$ Look for the coefficient of $y^3$ in the numerator and that would be $\binom{18}{3}=816$.