# Thread: proof with binomials

1. ## proof with binomials

Hi... my english is not the best (I speak spanish), but... I need help with some problems... Can you help me please? ^^... Is urgent!

Demonstrate:

2. Hi
Originally Posted by Mariposa
Hi... my english is not the best (I speak spanish), but...
That's not a problem
Demonstrate:

For the first one recall the binomial theorem : $(x+y)^n=\sum_{k=0}^n\binom{n}{k} x^ky^{n-k}$

As for any two integers $n$ and $k$ one has $1^k=1$ and $1^{n-k}=1$, one can write $\sum_{k=0}^n\binom{n}{k} = \sum_{k=0}^n\binom{n}{k}1^k1^{n-k}=(1+1)^n$ (thanks to the binomial theorem) and get $\sum_{k=0}^n\binom{n}{k}=2^n$. It gives us $\left[\sum_{k=0}^n\binom{n}{k}\right]^2=(2^n)^2=2^{2n}=(2^2)^n=4^n$

Using this idea, what is $\sum_{k=0}^n\binom{n}{k}3^k$ ?

For the last two questions use the definition of $\binom{n}{k}$ : $\binom{n}{k}=\frac{n!}{k!(n-k)!}$

For example : $\binom{n}{n-k}=\frac{n!}{(n-k)![n-(n-k)]!}=\ldots$

3. They were already solved in another forum.

4. $\sum\limits_{k = 0}^n {\left( {\begin{array}{*{20}c} n \\ k \\
\end{array}} \right)} 3^k = \left( {3 + 1} \right)^n = 4^n = \left( {2^n } \right)^2 = \left( {\sum\limits_{k = 0}^n {\left( {\begin{array}{*{20}c} n \\ k \\\end{array}} \right)} } \right)^2$