# Binomial Theorem?

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• Apr 28th 2009, 11:32 PM
fardeen_gen
Binomial Theorem?
Given that:
$\displaystyle S_{n} = 1 + q + q^2 + ... + q^n$
$\displaystyle \sigma_{n} = 1 + \left(\frac{q + 1}{2}\right) + \left(\frac{q + 1}{2}\right)^2 + ... + \left(\frac{q + 1}{2}\right)^n$
$\displaystyle q\neq1$
Prove that:
$\displaystyle {n + 1\choose 1} + {n + 1\choose 2}\cdot S_{1} + {n + 1\choose 3}\cdot S_{2} + ... + {n + 1\choose n + 1}\cdot S_{n} = 2^n\sigma_{n}$
• May 1st 2009, 03:34 PM
TheAbstractionist
Quote:

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

Hi fardeen_gen.

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

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

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

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

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

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

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

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

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

__$\displaystyle =\ 2^n\sigma_n$