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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

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Typo: your $+1$ should be $-1$. – Robert Israel Feb 13 '13 at 20:13
In the RHS of your equality you need minus signs both in the numerator and in the denominator...! – DonAntonio Feb 13 '13 at 20:13
@DonAntonio Care to copypaste from an answer you gave just now? – Git Gud Feb 13 '13 at 20:13
¿Have you tried to prove it by induction? – MyUserIsThis Feb 13 '13 at 20:14
Thanks @GitGud, I shall. – DonAntonio Feb 13 '13 at 20:14

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$$I=\frac{x(x^n-1)}{x-1}$$.

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Thank you very much! – Ignacio Ibarra Del Río Feb 14 '13 at 4:06

The following is the well known sum for a finite geometric sequence:


Please do note the indexes in both sums...

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Consider the following:


$S = \sum\limits_{i=1}^n(a+b)^i = (a+b)+(a+b)^2+(a+b)^3+\dots+(a+b)^n$


$(a+b)S =\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ (a+b)^2+(a+b)^3+\dots+(a+b)^n+(a+b)^{n+1}$

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, $$S = \dfrac{(a+b)\left((a+b)^n-1\right)}{a+b-1}$$

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Thank you very much for this answer. It was exactly what I wanted. Cheers! – Ignacio Ibarra Del Río Feb 14 '13 at 4:04
No problem :) Glad I was able to help you – Enzo Feb 14 '13 at 5:50

$$\sum^n_{i=1}{(a+b)^i} =\left(a+b\right) \cdot \frac{\left(a-b\right)^{n}-1}{a+b-1}.$$ Other words: geometric progression.

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