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proving $\displaystyle \binom{n}{0}-\binom{n}{1}+\binom{n}{2}+\cdots \cdots +(-1)^{\color{red}{m}-1}\binom{n}{m-1}=(-1)^{m-1}\binom{n-1}{m-1}.$

$\displaystyle \Rightarrow 1-n+\frac{n(n-1)}{2}+\cdots \cdots (-1)^{n-1}\frac{n.(n-1)\cdot (n-2)\cdots(n-m+2)}{(m-1)!}$

Added

writting LHS as $\displaystyle \binom{n}{0}-\left(\binom{n-1}{0}+\binom{n-1}{1}\right)+\left(\binom{n-1}{1}+\binom{n-1}{2}\right)+\cdots \cdots +(-1)^{n-1}\left(\binom{n-1}{m-2}+\binom{n-1}{m-1}\right)=(-1)^{m-1}\binom{n-1}{m-1}.$

$\displaystyle \binom{n}{0}-\binom{n-1}{0}+\binom{n-1}{1}-\cdots +(-1)^{n-1}\binom{n-1}{m-2}-\left(\binom{n-1}{1}-\binom{n-1}{2}+\cdots +(-1)^n\binom{m-1}{m-1}\right)$

wan,t be able to solve after that, help me to solve it

DXT
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2 Answers2

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Extended HINT: The result is incorrect as originally stated; it should read

$$\binom{n}0-\binom{n}1+\binom{n}2-\ldots+(-1)^{\color{crimson}m-1}\binom{n}{m-1}=(-1)^{m-1}\binom{n-1}{m-1}\;,$$

or, more compactly,

$$\sum_{k=0}^{m-1}(-1)^k\binom{n}k=(-1)^{m-1}\binom{n-1}{m-1}\;.\tag{1}$$

Fix $n\in\Bbb N$. For $m=1$ the desired result is

$$\binom{n}0=(-1)^0\binom{n-1}0\;,$$

which is indeed true, since both sides are equal to $1$. Suppose as an induction hypothesis that $(1)$ holds for some $m$; for the induction step we want to prove that

$$\sum_{k=0}^m(-1)^k\binom{n}k=(-1)^m\binom{n-1}m\;.\tag{2}$$

Using the induction hypothesis we can rewrite the lefthand side of $(2)$:

$$\sum_{k=0}^m(-1)^k\binom{n}k=(-1)^m\binom{n}m+\sum_{k=0}^{m-1}(-1)^k\binom{n}k=(-1)^m\binom{n}m+(-1)^{m-1}\binom{n-1}{m-1}\;,$$

so to complete the induction step we need only show that

$$(-1)^m\binom{n}m+(-1)^{m-1}\binom{n-1}{m-1}=(-1)^m\binom{n-1}m\;.$$

This is easily done using one of the most basic identities involving binomial coefficients.

Added: After some thought I realize that $(1)$ can be proved by direct calculation:

$$\begin{align*} \sum_{k=0}^{m-1}(-1)^k\binom{n}k&=\sum_{k=0}^{m-1}(-1)^k\left(\binom{n-1}k+\binom{n-1}{k-1}\right)\\ &=\sum_{k=0}^{m-1}(-1)^k\binom{n-1}k+\sum_{k=0}^{m-1}(-1)^k\binom{n-1}{k-1}\\ &=\sum_{k=0}^{m-1}(-1)^k\binom{n-1}k+\sum_{k=1}^{m-1}(-1)^k\binom{n-1}{k-1}\\ &=\sum_{k=0}^{m-1}(-1)^k\binom{n-1}k+\sum_{k=0}^{m-2}(-1)^{k+1}\binom{n-1}k\\ &=\sum_{k=0}^{m-1}(-1)^k\binom{n-1}k-\sum_{k=0}^{m-2}(-1)^k\binom{n-1}k\\ &=(-1)^{m-1}\binom{n-1}{m-1}+\sum_{k=0}^{m-2}(-1)^k\binom{n-1}k-\sum_{k=0}^{m-2}(-1)^k\binom{n-1}k\\ &=(-1)^{m-1}\binom{n-1}{m-1}\;. \end{align*}$$

Brian M. Scott
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  • To Brian M. Scott, please explain me without using induction, thanks – DXT Jan 02 '17 at 06:37
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    @Durgesh: The induction argument is very easy and almost certainly the simplest argument, but I’ll see if I can come up with some other argument. – Brian M. Scott Jan 02 '17 at 06:46
  • @Durgesh: I was wrong: there is a simpler argument. The result can be established by direct calculation, without induction. – Brian M. Scott Jan 02 '17 at 12:19
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HINT: Fix a random $n$ and do induction on $m$.

Stefan4024
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