Evaluate $1^{1337}\binom{n}{1} - 2^{1337}\binom{n}{2} + 3^{1337}\binom{n}{3} - 4^{1337}\binom{n}{4} + \cdots + (-1)^{n-1} n^{1337}\binom{n}{n}$ for a given integer $n>1337.$ Hint: Consider $(1+x)^n$.
1. Evaluate $\sum^{10}_{x=2}(ax+b)$ in terms of $a$ and $b.$
2. Since $\binom{n}{k}$ denotes the number of ways of choosing $k$ objects from $n$ objects, explain why $\binom{n+1}{k} = \binom{n}{k} + \binom{n}{k-1}.$

How would you use binomial expansion to expand $(1+x)^n?$

Could you use this expansion to evaluate $\binom{n}{1} - \binom{n}{2} + \binom{n}{3} - \binom{n}{4} + \cdots + (-1)^{n-1} \binom{n}{n}?$

How would you evaluate $1\binom{n}{1} - 2\binom{n}{2} + 3\binom{n}{3} - 4\binom{n}{4} + \cdots + (-1)^{n-1} n\binom{n}{n}?$ (Hint: Find a relationship between this expression and the equation in the first hint.)

Have you tried differentiating both sides of the equation in the first hint?

Could you use a similar method to evaluate $1^{2}\binom{n}{1} - 2^{2}\binom{n}{2} + 3^{2}\binom{n}{3} - 4^{2}\binom{n}{4} + \cdots + (-1)^{n-1} n^{2}\binom{n}{n}?$

… by first introducing another factor of $x$ to both sides of the equation achieved in the fourth hint.

Remember to use results you’ve already shown.

We begin by expanding $(1+x)^n=1+\binom{n}{1}x+\cdots+\binom{n}{n}x^n.$ We can differentiate both sides of the equation to get $$\frac{d}{dx} (1+x)^n = n(1+x)^{n-1} = {n\choose1} + 2{n\choose2}x + \cdots + n{n\choose n}x^{n-1}$$ We can further increment the exponent on the constant factors by multiplying the whole thing by $x$ and differentiating again: \begin{aligned} nx(1+x)^{n-1} &= {n\choose1}x + 2{n\choose2}x^2 + \cdots + n{n\choose n}x^{n} \\ \frac{d}{dx} nx(1+x)^{n-1} = n(1+x)^{n-2}(nx + 1) &= {n\choose1} + 2^2{n\choose2}x + \cdots + n^2{n\choose n}x^{n-1} \\ \end{aligned} Since we have the condition that $n > 1337,$ we can repeat the process of multiplying by $x$ and differentiating, until we reach the desired exponent of $1337.$ Setting $x = -1$ gives us the desired alternating signs on the RHS expression, and since the LHS will still contain a factor of $(1+x),$ we get that the answer is $0.$