Using Newton's binomial theorem to prove that a sum evaluates to $36^n-26^n$

Question

Using Newton's binomial theorem to argue that: $n \ge 1$ $$36^n - 26^n = \sum_{k=1}^{n}\binom{n}{k}10^k \cdot 26^{n-k}$$ my argument $$(26+10)^n = \sum_{k=1}^{n}\binom{n}{k}10^k \cdot 26^{n-k} +26^n $$ $$(26+10)^n = \sum_{k=0}^{n}\binom{n}{k}10^k \cdot 26^{n-k} +26^n $$

I'm stuck at the part on what to do with the $26^n$ After I moved it over a made the sum $\sum_{k=1}^{n}$ to $\sum_{k=1}^{n}$

Answer 1

We first take the binomial expansion of $(26+10)^n$. $$ \begin{align} (26+10)^n&=\sum_{k=0}^n10^k\cdot26^{n-k}\\ &=26^n+\sum_{k=1}^n10^k\cdot26^{n-k} \end{align} $$ Notice that we pulled out $26^n$ from our sum.

Therefore, by subtracting $26^n$ from both sides we obtain $$\begin{align} (26+10)^n-26^n&=\sum_{k=1}^n10^k\cdot26^{n-k}\\ 36^n-26^n&= \end{align} $$

Answer 2

$$ \sum_{k=0}^n \binom n k 10^k \cdot 26^{n-k} = \underbrace{\sum_{k=0}^n \binom n k a^k b^{n-k} = (a+b)^n}_{\text{This is what the theorem says.}} = (10 + 26)^n = 36^n. $$

(You wanted to prove $$36^n-26^n = \sum_{k=1}^n\binom n k 10^k\cdot26^{n-k},$$ with the sum starting at $1$ rather than at $0$. If you add $26^n$ to both sides, you have $$36^n = 26^n + \sum_{k=1}^n\binom n k 10^k\cdot26^{n-k},$$ and that is the same as $$\binom n 0 10^0\cdot 26^{n-0} + \sum_{k=1}^n\binom n k 10^k\cdot26^{n-k} = \sum_{k=0}^n\binom n k 10^k\cdot26^{n-k},$$ with the sum starting at $0$, not at $1$, because the term $26^n$ is $\binom n 0 10^0\cdot 26^{n-0}$.)