5
$\begingroup$

Calculate the sum:

$ \sum_{k=0}^n (-1)^k {n+1\choose k+1} $

I don't know if I'm so tired or what, but I can't calculate this sum. The result is supposed to be $1$ but I always get something else...

  • 0
    possible duplicate of [Evaluate $ \bi$n$om{n}{$0$}+\bi$n$om{$n$}{2}+\bi$n$om{$n$}{4}+\cdots+\bi$n$om{$n$}{2k}+\cdots$](http://math.stackexcha$n$ge.com/questions/42797/evaluate-binomn0-binomn2-binomn4-cdots-binomn2k-cdots)2014-01-09

4 Answers 4

1

$\sum_{k=0}^n (-1)^k {n+1\choose k+1}=-\sum_{k=0}^n (-1)^{k+1} {n+1\choose k+1}= $ $=-\left(\sum_{k=0}^n (-1)^{k+1} {n+1\choose k+1}\right)=$ $=-\left(\sum_{j=1}^{n+1} (-1)^{j} {n+1\choose j}+(-1)^{0} {n+1\choose 0}-(-1)^{0} {n+1\choose 0}\right)= $ $=-\left(\sum_{j=0}^{n+1} (-1)^{j} {n+1\choose j}-(-1)^{0} {n+1\choose 0}\right)=-(1-1)^{n+1}+1=1$

5

HINT

  1. $\displaystyle (1-1)^{n+1} = \sum_{k = 0}^{n+1} {n \choose k}1^k(-1)^{n-k}$
  2. What terms are or aren't in your sum that are in the one above?
4

Using the binomial theorem we have: $ (1 + (-1))^{n+1} = {{n+1} \choose 0} (-1)^0 + {{n+1} \choose 1} (-1)^1 + \ldots + {{n+1} \choose {n+1}} (-1)^{n+1}.$ "Divide" by ${-1}$ to get: $ - (1 - 1)^{n+1} = -{{n+1} \choose 0} + \color{blue}{{{n+1} \choose 1} (-1)^0 + \ldots + {{n+1} \choose {n+1}} (-1)^{n}}.$ This pretty much solves it.

2

Another way to see it: prove that $\binom{n}{k}+\binom{n}{k+1}=\binom{n+1}{k+1}\,\,\,,\,\,\text{so}$ $\sum_{k=0}^n(-1)^k\binom{n+1}{k+1}=\sum_{k=0}^n(-1)^k\cdot 1^{n-k}\binom{n}{k}+\sum_{k=0}^n(-1)^k\cdot 1^{n-k}\binom{n}{k+1}=$ $=(1+(-1))^n-\sum_{k=0}^n(-1)^{k+1}\cdot 1^{n-k-1}\binom{n}{k+1}=0-(1-1)^n+1=1$