# Thread: Question Discrete Math Binomial Theorem

1. ## Question Discrete Math Binomial Theorem Please could someone help me the way to approach this?

2. ## Re: Question Discrete Math Binomial Theorem

The binomial theorem states (as you could have looked up)

$(a+b)^n = \sum \limits_{k=0}^n ~a^k ~b^{n-k}$

you should be able to figure out where to plug in $2$ and $(-1)$ into the formula above

3. ## Re: Question Discrete Math Binomial Theorem Originally Posted by romsek The binomial theorem states (as you could have looked up)

$(a+b)^n = \sum \limits_{k=0}^n ~a^k ~b^{n-k}$

you should be able to figure out where to plug in $2$ and $(-1)$ into the formula above
Not quite! You left out the binomial coefficient that was already in inayat's formula.

The binomial theorem says that

$(a+b)^n = \sum \limits_{k=0}^n \begin{pmatrix}n \\ k\end{pmatrix} ~a^k ~b^{n-k}$

Where $\begin{pmatrix}n \\ k\end{pmatrix}$ is the "binomial coefficient", $\begin{pmatrix}n \\ k\end{pmatrix}= \frac{n!}{k!(n-k)!}$.

4. ## Re: Question Discrete Math Binomial Theorem Originally Posted by inayat  Please could someone help me the way to approach this?
I hope the is a clear proof for you.
$\begin{gathered} \sum\limits_{k = 0}^n {\dbinom{n}{k}{a^{n - k}}{b^k} = {{(a + b)}^n}} \hfill \\ \sum\limits_{k = 0}^n {\dbinom{n}{k}{2^{n - k}}{{( - 1)}^k} = {{(2 + ( - 1))}^n}} \hfill \\ = 1 \hfill \\ \end{gathered}$