$1^{1337}\binom{n}{1}  2^{1337}\binom{n}{2} + 3^{1337}\binom{n}{3}  4^{1337}\binom{n}{4} + \cdots + (1)^{n1} n^{1337}\binom{n}{n}$
for a given integer $n>1337.$
Hint: Consider $(1+x)^n$
.

 Evaluate
$\sum^{10}_{x=2}(ax+b)$
in terms of$a$
and$b.$
 Since
$\binom{n}{k}$
denotes the number of ways of choosing$k$
objects from$n$
objects, explain why$\binom{n+1}{k} = \binom{n}{k} + \binom{n}{k1}.$
 Evaluate

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

Could you use this expansion to evaluate
$\binom{n}{1}  \binom{n}{2} + \binom{n}{3}  \binom{n}{4} + \cdots + (1)^{n1} \binom{n}{n}?$

How would you evaluate
$1\binom{n}{1}  2\binom{n}{2} + 3\binom{n}{3}  4\binom{n}{4} + \cdots + (1)^{n1} n\binom{n}{n}?$
(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
$1^{2}\binom{n}{1}  2^{2}\binom{n}{2} + 3^{2}\binom{n}{3}  4^{2}\binom{n}{4} + \cdots + (1)^{n1} n^{2}\binom{n}{n}?$

… 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$$ \frac{d}{dx} (1+x)^n = n(1+x)^{n1} = {n\choose1} + 2{n\choose2}x + \cdots + n{n\choose n}x^{n1} $$
We can further increment the exponent on the constant factors by multiplying the whole thing by$x$
and differentiating again:$$\begin{aligned} nx(1+x)^{n1} &= {n\choose1}x + 2{n\choose2}x^2 + \cdots + n{n\choose n}x^{n} \\ \frac{d}{dx} nx(1+x)^{n1} = n(1+x)^{n2}(nx + 1) &= {n\choose1} + 2^2{n\choose2}x + \cdots + n^2{n\choose n}x^{n1} \\ \end{aligned}$$
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$0.$