# Binomial Theorem identities, evaluate the sum

This is a homework problem, please don't blurt out the answer! :)

I've been given the following, and asked to evaluate the sum:

$$\sum_{k = 0}^{n}(-1)^k\binom{n}{k}10^k$$

So, I started out trying to look at this as equivalent to the binomial theorem, in which case, I could attempt something like this: $10^k = y^{n-k}$ but I didn't feel that got me anywhere.

So I started actually evaluating it...

$$(-1)^0\binom{n}{0}10^0 + (-1)^1\binom{n}{1}10^1 + \ldots + (-1)^n\binom{n}{n}10^n$$

So, if I'm thinking correctly, all the other terms cancel out and you are left with:

$$(-1)^n\binom{n}{n}10^n = (-1)^n10^n$$

But, obviously this cannot be correct (or can it?). The book gives a slightly different answer, so I'm wondering where I'm going wrong. Some direction would be greatly appreciated!

Books answer: $\displaystyle (-1)^n9^n$

-

Try to fit your sum into one of the following: $$\sum_{k=0}^n\binom{n}ka^kb^{n-k}=(a+b)^n,\quad\sum_{k=0}^n\binom{n}kb^{n-k}=(1+b)^n,\quad\sum_{k=0}^n\binom{n}ka^k=(a+1)^n.$$

-
I can see how it fits into the last one but for example, in the first sum, how does supplying 0 for b equate to it being (a+1) in the last sum? why not (a+0)? –  intervade Jan 26 '13 at 20:37
Plugging in $b=0$ kills every term except $a^n$. What you want is to have a sum of powers of $a$, from $1=a^0$ to $a^n$, and for that you need the factor $b^{n-k}$ to be equal to $1$ for every $k$, that is, $b=1$. –  Did Jan 26 '13 at 20:41

Hint: If you're evaluating $(a+b)^n$, what does the binomial theorem say this is equal to and how do you relate that to the sum you're given?

-

From binomial theorem$$\sum_{k = 0}^{n}\binom{n}{k}t^k=(1+t)^k$$ for $t=-10$ $$\sum_{k = 0}^{n}\binom{n}{k}(-10)^k=(1-10)^n$$ $$\sum_{k = 0}^{n}(-1)^k\binom{n}{k}10^k=\sum_{k = 0}^{n}\binom{n}{k}(-10)^k=(1-10)^n=(-9)^n$$

-

$\displaystyle \sum_{k=0}^n\binom{n}ka^kb^{n-k}=(a+b)^n$

here $a=10$.

if $n$ is even then $b=-1$ we have : $\displaystyle \sum_{k=0}^n\binom{n}{k}10^k(-1)^{n-k}=\sum_{n=0}^{2m}\binom{2m}{k}(-1)^k 10^k=(10-1)^n=9^n$

if $n$ is odd : $\displaystyle \sum_{k=0}^n\binom{n}{k}10^k(-1)^{n-k}=-\sum_{n=0}^{2m+1}\binom{2m+1}{k}(-1)^k 10^k=-(10-1)^n=-9^n$

hence, you get the answer in the book.

-