# Proof using Binomial Theorem

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• May 29th 2011, 04:09 PM
Oiler
Proof using Binomial Theorem
Hey all,

I am stuck on trying to prove $2^n = \sum_{r=0}^r \begin{pmatrix} n\\ r \end{pmatrix}$ using the Binomial Theorem, any ideas ? Thanks
• May 29th 2011, 04:15 PM
Also sprach Zarathustra
Quote:

Originally Posted by Oiler
Hey all,

I am stuck on trying to prove $2^n = \sum_{r=0}^r \begin{pmatrix} n\\ r \end{pmatrix}$ using the Binomial Theorem, any ideas ? Thanks

Hint:

2=1+1 (In most of the times)
• May 29th 2011, 04:22 PM
Plato
Quote:

Originally Posted by Oiler
$2^n = \sum_{r=0}^r \begin{pmatrix} n\\ r \end{pmatrix}$ using the Binomial Theorem

This is a standard theorem if we know that $\left( {a + b} \right)^n = \sum\limits_{k = 0}^n {\binom{n}{k}a^k b^{n - k} }$
Now let $a=1~\&~b=1$

NOTE you have a mistake. The upper limit in n.