AIME 1986 problem

The given equation is : $1-x+x^2-x^3+\cdots -x^{17}$ It is easy to see that it is a GP of $18$ terms with first term $1$ and common ratio $-x$ :

So the above sum can be rewritten as: $\dfrac{1(1-(-x))^{18}}{1-(-x)}$ $=\dfrac{1-x^{18}}{1+x}$ Now we can replace $x$ with $y-1$ : Hence the sum is rewritten as : $\dfrac{1-(y-1)^{18}}{y}$ Since $y-1$ is in the even power hence the term inside can be inverted to $1-y$ : Hence the sum becomes: $\dfrac{1-(1-y)^{18}}{y}$ Expanding $(1-y)^{18}$ using binomial theorem: $\dfrac{1-1+\binom{18}{1}y-\binom{18}{2}y^2+\binom{18}{3}y^3-\cdots-\binom{18}{18}y^{18}}{y}$ $=\binom{18}{1}-\binom{18}{2}y+\binom{18}{3}y^2+\cdots-\binom{18}{18}y^{17}$ Hence the required answer is $\boxed{\binom{18}{3}=816.}$

Let $\quad \displaystyle f(x) = \sum_{n=0}^{17} {(-x)^n} = g(x) = \sum_{n=0}^{17} {a_ny^n}$ .

Then $\quad \dfrac {df(x)}{dx} = \dfrac {dg(y)}{dx} = \dfrac {dg(y)}{dy} \dfrac {dy}{dx} = \dfrac {dg(y)}{dy}$ .

Similarly, $\quad \dfrac {d^2f(x)}{dx^2} = \dfrac {d^2g(y)}{dy^2}$

We note that $\displaystyle \quad \dfrac {df(x)}{dx} = \sum_{n=1}^{17} {[(-1)^nnx^{n-1}]} \quad$ and $\displaystyle \quad \dfrac {d^2f(x)}{dx^2} = \sum_{n=2}^{17} {[(-1)^nn(n-1)x^{n-2}]}$

We also know that:

$g''(0) = \left[ \dfrac {d^2g(y)}{dy^2} \right]_{y=0} = 2a_2 = f''(-1) = \left[ \dfrac {d^2f(x)}{dx^2} \right]_{x=-1}$

$= \displaystyle \sum_{n=2}^{17} {n(n-1)} = \sum_{n=1}^{17} {n^2} - \sum_{n=1}^{17} {n} = \dfrac {(17)(18)(35)}{6}-\dfrac {(17)(18)}{2} = 1632$

$\Rightarrow a_2 = \dfrac {1632}{2} = \boxed{816}$

Tried SUM 1 to 5 and found that for SUM 1 to odd n, A(o) = A(n-1), A(1) = A(n-2),, A(2) = A(n-3), etc
Coff. of X^k term in both F(X) and G(Y) must be the same. coff. of X^17 is -1 in F. Hence A(17) = -1
F(-1) = 18, but F(-1) = G(o) = A(o) + o + o + . . . + o...=A(n-1) =A(17 - 1) = A(16)........................A(16) = 18

Expanding (X + 1)^k by Binomial Theorem we get the coff. of X^15 and X^14 for k= 14 to 17.
.........................(X + 1)^14.................(X + 1)^15.....................(X + 1)^16.....................(X + 1)^17
FOR X^15................................................1..........................................16.............................136
for G..................................................1 A(15)..........................16 A(16).....................136 A(17) .....
for G....................................................A(15)..........................16 (18)........................136*(-1) =-1 from F .................................................................................................................................................................A(15)= - 153.

FOR X^15.............1.................................15...............................120..............................680
for G..................1 A(14)........................15 A(15)..................120 A(16).....................680 A(17) .....
for G.................... A(14)..........................15 (-153).................120 (18)........................680*(-1) =-1 from F
..................................................................................................................................A(2) = A(14) =816

Using the geometric series formula, 1 - x + x^2 + \cdots - x^{17} = \frac {1 - x^{18}}{1 + x} = \frac {1-x^{18}}{y}. Since x = y - 1, this becomes \frac {1-(y - 1)^{18}}{y}. We want a_2, which is the coefficient of the y^3 term in -(y - 1)^{18} (because the y in the denominator reduces the degrees in the numerator by 1). By the binomial theorem that is (-1) \cdot (-1)^{15}{18 \choose 3} = \boxed{816}.