2
$\begingroup$

Find the value of $\displaystyle 99^{50}-\binom{99}{1}(98)^{50}+\binom{99}{2}(97)^{50}-\cdots \cdots +99$

Binomial identity:

$\displaystyle (1-x)^{99} = \binom{99}{0}-\binom{99}{1}x+\binom{99}{2}x^2-\cdots \cdots -\binom{99}{99}x^{99}$

I want be able to go further, could some help me with this, thanks

  • 0
    the $-$ and $+$ are alternating or not? – 2017-01-10
  • 0
    yes @Exodd - and + are alternate. – 2017-01-10
  • 0
    If your signs are alternating, this looks like an [inclusion-exclusion](https://en.wikipedia.org/wiki/Inclusion%E2%80%93exclusion_principle) thing -- but it's a bit strange because it seems to try to compute "the number of length-50 sequences drawn from 99 symbols that don't miss any of the 99 symbols", which is impossible. – 2017-01-10
  • 1
    @Rohan -- to fit the pattern it should be $\binom{99}{98}1^{50}=99$. – 2017-01-10
  • 0
    @Rohan: It's easier to write the general term as $(-1)^{n+1}\binom{99}{n} n^{50}$, with $n$ _decreasing_ from $99$ to $1$. Then the last term is $\binom{99}{1} 1^{50} = 99$, as written.. – 2017-01-10
  • 0
    @HenningMakholm yes, that's what it computes. Since such a sequence is impossible, the answer must be 0. – 2017-01-10

1 Answers 1

2

It is simply zero. Let $\delta$ be the (difference) operator mapping $p(x)$ into $p(x)-p(x-1)$: if $p(x)$ is a polynomial with degree $d\geq 1$, then $(\delta p)(x)$ is a polinomial with degree $d-1$. Since

$$(\delta^n p)(x) = \sum_{k=0}^{n}(-1)^k \binom{n}{k} p(x-k) \tag{1}$$ if $p(x)$ is a polynomial with degree $ The given sum is the RHS of $(1)$ in the case $p(x)=x^{50}$, $n=99=x$.