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$$\binom{n}{1}-2\binom{n}{2}+3\binom{n}{3}+\cdots+(-1)^{n-1}n\binom{n}{n}=0$$

I tried solving this identity using this one: $$(n+1) \cdot C(n,k) = (k+1) \cdot C(n+1,k+1)$$ but I didn't get any far.

Any hints for the solution would be appreciated :)

Thanks in advance :)

Blue
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mathematics
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2 Answers2

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Assuming that you are familiar with differentiation, expand $(1-x)^n$ using binomial theorem and then differentiate, finally substitute $1$ in the expression.

We can do like this $$(1-x)^n = \sum_{k = 0}^{n} C(n,k) (-1)^k x^k$$

Now differentiating both sides, $$-n (1-x)^{n-1} = \sum_{k= 1}^{n} kC(n,k) (-1)^k x^{k-1}$$ and substitute $1$

Infinity_hunter
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By binomial identity \begin{align} (1 - x)^n = 1 - \binom{n}{1}x + \binom{n}{2}x^2 + \cdots + (-1)^n\binom{n}{n}x^n. \end{align} Taking derivative with respect to $x$ on both sides yields \begin{align*} n(1 - x)^{n - 1}(-1) = -\binom{n}{1} + 2\binom{n}{2}x + \cdots + (-1)^nn\binom{n}{n}x^{n - 1}. \end{align*} Setting $x = 1$ on the above equation then gives the desired result.

Zhanxiong
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