$1^{1337}\binom{n}{1} - 2^{1337}\binom{n}{2} + 3^{1337}\binom{n}{3} - 4^{1337}\binom{n}{4} + \cdots + (-1)^{n-1} n^{1337}\binom{n}{n}$ for a given integer $n>1337.$
Hint: Consider $(1+x)^n$.
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- 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$\binom{n+1}{k} = \binom{n}{k} + \binom{n}{k-1}.$
- Evaluate
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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?
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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.
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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$0.$