# combinatorics

• May 1st 2010, 05:38 PM
sofialam
combinatorics
• May 1st 2010, 10:53 PM
hollywood
$\sum_{i=0}^N\left(\begin{array}{c}N\\i\end{array}\ right)a^{i+1}b^{N-i}$

You shift the index of summation - instead of i running from 0 to N, we want it to run from 1 to N+1, so replace i with i-1:

$\sum_{i-1=0}^{i-1=N}\left(\begin{array}{c}N\\i-1\end{array}\right)a^{i-1+1}b^{N-(i-1)}$

and simplify:

$\sum_{i=1}^{N+1}\left(\begin{array}{c}N\\i-1\end{array}\right)a^ib^{N+1-i}$

Now bring out the N+1 term:

$\left(\begin{array}{c}N\\(N+1)-1\end{array}\right)a^{N+1}b^{N+1-(N+1)}+\sum_{i=1}^N\left(\begin{array}{c}N\\i-1\end{array}\right)a^ib^{N+1-i}$

which simplifies to:

$a^{N+1}+\sum_{i=1}^N\left(\begin{array}{c}N\\i-1\end{array}\right)a^ib^{N+1-i}$

- Hollywood