Simplifying this sigma notation

Question

Is there any way I can simplify this sigma notation?

$$\begin{align*} \sum_{k=1}^m(5^k) \end{align*}$$

Answer 1

Yes, there is.

Let $S = \displaystyle\sum_{k=1}^{m} 5^{k}$ be the sum you're trying to simplify, then you'll have $$5S=\color{red}{5^2+5^3+\cdots+5^m}+5^{m+1}$$ $$S=5^1+\color{red}{5^2+\cdots+5^{m-1}+5^m}$$ Then, substracting $S$ from $5S$ gives you $$4S=5^{m+1}-5^1\iff S=\frac{5^{m+1}-5}{4}.$$

In general, for $a\not =1$, you'll get the following using the above method : $$\sum_{k=1}^{m}a^k=\frac{a^{m+1}-a}{a-1}.$$ Note that if $a=1$, then $\sum_{k=1}^{m}1^k=m$ trivially.

Answer 2

$5+5^2+5^3+...+5^m=5\large\frac{(5^m-1)}{(5-1)}$

It is a GP with first term=$5$ and common ration=$5$

Answer 3

Using the formula $$\sum_{k=0}^{N} x^k=\frac{x^{N+1}-1}{x-1}$$ we have:

$$\sum_{k=1}^m 5^k=\sum_{k=1}^m 5^k+5^0-5^0=\sum_{k=0}^m 5^k-1=\frac{5^{m+1}-1}{5-1}-1=\frac{5^{m+1}-1}{4}-1=\frac{5^{m+1}-1-4}{4}=\frac{5^{m+1}-5}{4}$$