These sums are getting out of hand

As generating functions, $(\lambda_0 + \lambda_1 x + \lambda_2 x^2 + \dotsb)^3 = 1 + x + x^2 + \dotsb = (1 - x)^{-1},$ so $\lambda_0 + \lambda_1 x + \lambda_2 x^2 + \dotsb = (1 - x)^{-\frac{1}{3}}.$ Taking the coefficient of $x^n$ in both sides, we get $\lambda_n = (-1)^n \binom{-\frac{1}{3}}{n}.$

Note: As stated, the problem is ambiguous, since $-\frac{1}{3}$ can be written as $\frac{-1}{3}$ or $\frac{1}{-3}$ .

Using $n = 0$ , we get $\lambda_0^3 = 1$ or $\lambda_0 = 1$ .

Using $n = 1$ , we get $3 \lambda_0^2 \lambda_1 = 1$ or $\lambda_1 = \frac{1}{3}$ .

Matching this to the given form, we have:

$\displaystyle\begin{aligned} \lambda_1 &= (-1)^1 \binom{p/q}{1} \\ \frac{1}{3} &= -\frac{p}{q} \\ \frac{p}{q} &= \frac{-1}{3} \end{aligned}$

Thus $p = -1, q = 3$ , and $p+q = \boxed{2}$ .

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Tip: Instead of asking for the general form (and thus can be abused like above), ask for some specific value like $\lambda_{2015}$ or something.

EDIT: This solution was written when the question said $\lambda_n = (-1)^n \binom{p/q}{n}$ for all $n$ , and finding $p+q$ .