# Binomial Theorem?

• Apr 29th 2009, 12:32 AM
fardeen_gen
Binomial Theorem?
Given that:
$S_{n} = 1 + q + q^2 + ... + q^n$
$\sigma_{n} = 1 + \left(\frac{q + 1}{2}\right) + \left(\frac{q + 1}{2}\right)^2 + ... + \left(\frac{q + 1}{2}\right)^n$
$q\neq1$
Prove that:
${n + 1\choose 1} + {n + 1\choose 2}\cdot S_{1} + {n + 1\choose 3}\cdot S_{2} + ... + {n + 1\choose n + 1}\cdot S_{n} = 2^n\sigma_{n}$
• May 1st 2009, 04:34 PM
TheAbstractionist
Quote:

Originally Posted by fardeen_gen
Given that:
$S_{n} = 1 + q + q^2 + ... + q^n$
$\sigma_{n} = 1 + \left(\frac{q + 1}{2}\right) + \left(\frac{q + 1}{2}\right)^2 + ... + \left(\frac{q + 1}{2}\right)^n$
$q\neq1$
Prove that:
${n + 1\choose 1} + {n + 1\choose 2}\cdot S_{1} + {n + 1\choose 3}\cdot S_{2} + ... + {n + 1\choose n + 1}\cdot S_{n} = 2^n\sigma_{n}$

Hi fardeen_gen.

Note that $S_r=\frac{1-q^{r+1}}{1-q},\quad0\le r\le n.$

Hence $\sum_{r\,=\,0}^n{n+1\choose r+1}S_r$

__ $=\ \frac1{1-q}\sum_{r\,=\,0}^n{n+1\choose r+1}(1-q^{r+1})$

__ $=\ \frac1{1-q}\sum_{r\,=\,0}^n{n+1\choose r+1}\,-\,\frac1{1-q}\sum_{r\,=\,0}^n{n+1\choose r+1}q^{r+1}$

__ $=\ \frac1{1-q}\left[2^{n+1}-1\right]\,-\,\frac1{1-q}\left[(1+q)^{n+1}-1\right]$

__ $=\ \frac{2^{n+1}-(1+q)^{n+1}}{1-q}$

__ $=\ \frac{2^{n+1}\left[1-\left(\frac{1+q}2\right)^{n+1}\right]}{1-q}$

__ $=\ \frac{2^n\left[1-\left(\frac{1+q}2\right)^{n+1}\right]}{\frac{1-q}2}$

__ $=\ \frac{2^n\left[1-\left(\frac{1+q}2\right)^{n+1}\right]}{1-\frac{1+q}2}$

__ $=\ 2^n\sigma_n$