I'm practicing for my maths term test mainly on binomial coefficients. I can't seem to find out how to prove the following identity. Any advice?
$$ \sum\limits_{k=1}^n (-1)^{k+1} k{{n}\choose k} = 0 $$
Thanks in advance.
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1usually instead of $x$ you need to use $k$ or $i$ $x$ is reserved for real numbers – Adi Dani May 22 '13 at 18:13
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Hint: Note that for $x\ge 1$, $x\dbinom{n}{x}=n\dbinom{n-1}{x-1}$. This can be verified by using the representation of the binomial coefficients in terms of factorials.
We can bring the $n$ to the front, and end up with something that is perhaps familiar. If it is not familiar, expand $(1-1)^m$ using the Binomial Theorem.
André Nicolas
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1Yup,then it becomes the product of $n$ and the difference of the sum of coefficients with even indexes and odd, respectively, which is null. Btw. I skipped over the previous problem which asked me to prove what you just told me. xD – nikitautiu May 22 '13 at 18:01
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$$\sum_{k=1}^n(-1)^{k+1}k\binom{n}{k}=\sum_{k=1}^n(-1)^{k+1}k\frac{n}{k}\binom{n-1}{k-1}= $$ $$=n\sum_{k=1}^n(-1)^{k+1}\binom{n-1}{k-1}=n\sum_{k=0}^{n-1}(-1)^{k+2}\binom{n-1}{k}=$$ $$n\sum_{k=0}^{n-1}(-1)^{k}\binom{n-1}{k}=n\sum_{k=0}^{n-1}\binom{n-1}{k}(-1)^{k}=$$ $$n(1+(-1))^{n-1}=n\cdot 0=0$$ From binomial theorem $$(1+x)^{n-1}=\sum_{k=0}^{n-1}\binom{n-1}{k}x^k$$ for $x=-1$ we get that $$(1+(-1))^{n-1}=\sum_{k=0}^{n-1}\binom{n-1}{k}(-1)^{k}=0$$
Adi Dani
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