Wolfram and a differential equations book that I read give me this form as solution
$$\ \sum^n_{i=1}{(a+b)^i} = \frac{(a+b)((a+b)^n-1)}{(a+b-1)}$$
However, I would like to get there step by step, in order to understand the logic between expression and solution.
Thanks for any advice or clue, Ignacio
This sum is called a finite geometric progression and we have$$\displaystyle\sum_{i=1}^{n}x^i=\frac{x(x^{n}-1)}{x-1}\hspace 10mm(*)$$ in your case $x=a+b$ which implies that$$\displaystyle\sum_{i=1}^n(a+b)^i=\frac{(a+b)((a+b)^{n}-1)}{a+b-1}$$If you like to know why (*) is true we have the following$$I=\displaystyle\sum_{i=1}^{n}x^i=x+x^2+...+x^n$$ but then $$xI=x^2+x^3+...+x^{n+1}$$ which implies that$$xI-I=(x^2+x^3+...+x^{n+1})-(x+x^2+...+x^n)$$but then we have$$I(x-1)=x^{n+1}-x=x(x^n-1)$$Which implies that$$I=\frac{x(x^n-1)}{x-1}$$.
The following is the well known sum for a finite geometric sequence:
$$\sum_{k=1}^nx^{k-1}=\sum_{k=0}^{n-1}x^k=\frac{1-x^{n-1}}{1-x}$$
Please do note the indexes in both sums...
Consider the following:
Let
$S = \sum\limits_{i=1}^n(a+b)^i = (a+b)+(a+b)^2+(a+b)^3+\dots+(a+b)^n$
Thus
$(a+b)S =\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ (a+b)^2+(a+b)^3+\dots+(a+b)^n+(a+b)^{n+1}$
Pay attention to how I aligned the two equations.
Now subtract $S$ from $(a+b)S$, it should be quite obvious that
$$(a+b)S-S = (a+b)^{n+1}-(a+b)$$ $$(a+b-1)S = (a+b)\left((a+b)^n-1\right)$$
Hence, $$S = \dfrac{(a+b)\left((a+b)^n-1\right)}{a+b-1}$$
$$\sum^n_{i=1}{(a+b)^i} =\left(a+b\right) \cdot \frac{\left(a-b\right)^{n}-1}{a+b-1}.$$ Other words: geometric progression.
