# Prove this equality by using Newton's Binomial Theorem

Let $n \ge 1$ be an integer. Use newton's Binomial Theorem to argue that

$$36^n -26^n = \sum_{k=1}^{n}\binom{n}{k}10^k\cdot26^{n-k}$$

I do not know how to make the LHS = RHS. I have tried $(36^n-26^n) = 10^n$ which is $x$ in the RHS, but I don't know what to do with the $26^{n-k}$ after I have gotten rid of the $26^n$ on the right. I also know I might have to use $\binom{n}{k} = \binom{n-1}{k-1} + \binom{n-1}{k}$ pascal's identity in this question.

Maybe I am approaching it from a completely wrong point of view, If someone can help point me in the right direction. It would be much appreciated!!!

• Rolled back destructive edit – apnorton Oct 1 '14 at 0:02

Bring the $26^n$ to the other side. You are then looking at the binomial expansion of $(26+10)^n$. The $26^n$ is the $k=0$ term that was missing in the given right-hand side.
• at which point the sum on the RHS starts at $k=0$ rather than $k=1$ – Henry Sep 27 '14 at 23:38
• It is the usual $(a+b)^n=\sum_0^n \binom{n}{k}a^k b^{n-k}$. – André Nicolas Sep 28 '14 at 6:06