Sum of geometric progression of $(1+x)^{m+n}$

Question

$$\displaystyle \left[(1+x)^m+(1+x)^{m+1}+..........+(1+x)^{m+n}\right] = \frac{(1+x)^m((1+x)^{n+1}-1)}{(1+x)-1} = \frac{(1+x)^{m+n+1}-(1+x)^{m}}{x}$$

I have the above sum of geometric progression, but I don't exactly see how it has been derived, I guess some of the algebraic steps has been hidden. Because, as long as I know, sum of geometric progression has the following formula: $$\sum_{k=1}^{n} ar^{k-1} = \frac{a(1-r^n)}{1-r}$$

But, surely it is using another related formula of this one. If someone can give me a detailed algebraic sum of geometric progression for the above sum, I would be glad.

Answer 1

It is exactly the formula you are mentioning, with $a=(1+x)^m$, $r=1+x$, and $n+1$ instead of $n$.

Answer 2

Very simple: $$\sum_{k=m}^{m+n}a^k=a^m\sum_{k=0}^na^k=a^m\frac{a^{n+1}-1}{a-1}.$$