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I found this equation in a computer science article: $$ \sum^{m}_{i=0} N^{i} = \frac{N^{m+1}-1}{N-1} $$ and I'm really unsure why this is true. I tried to come up with an answer myself, I arbitrarily chose $m$ to be equal to 4, here are my equations: $$ \sum^{4}_{i=0} N^{i} = 1 + N + N^{2} + N^{3} + N^{4} = N(N^{3} + N^{2} + N + 1) + 1 $$ After this I'm not sure where to go, somehow I should be able to reach $N^{5}$ (as in $\frac{N^{5}-1}{N-1}$) but I don't see how that's possible.

Blue
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Patyo21
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2 Answers2

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$$\color{red}{S_m} = \sum_{i = 0}^m {N^i} = \color{blue}{1} + \color{red}{N + N^2 +..... N^{m}}$$ $$\color{green}{N\times S_m} =N\times \sum_{i = 0}^m {N^i} = \sum_{i = 0}^m N^{i+1} = \color{red}{N + N^2 + N^3 +...N^{m}}+ \color{blue}{N^{m+1}} $$ Subtraction $S_m \text{ from }N\times S_m$ which will cancel out the red colored terms and only blue colored terms will left

On the left side we have, $N\times S_m -S_m = S_m(N-1)$ while on the right side we'll have $(N^{m+1}-1)$

Thus, $$\color{blue}{N\times S_m}-\color{blue}{S_m} = \color{brown}{S_m}\times \color{green}{(N-1)} = \color{green}{N^{m+1} - 1} \text{ dividing both sides with }(N-1)\\\implies \color{brown}{S_m} = \frac {N^{m+1}-1}{N-1}$$

For the above solution, we can divide with $(N-1)$ only when $N\ne1$ but we still can use the above formula as $\lim_{N\to1}$ and apply L'Hôpital's rule

Darshan P.
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    I almost understand your solution, I only have two further questions, why does this equation apply: $S m \times (N - 1) = N^{m+1} - 1$ ? I understand that $Sm \times N$ is equal to $N^{m+1}$, but why is $Sm \times (-1)$ equal to $(-1)$?

    My other question is why does $N \times Sm - Sm = N^{m+1} - 1$ imply $S = \frac{N^{m+1} - 1}{N - 1}$? I don't see the connection there, I'm kind of inexperienced in this field of maths.

     – Patyo21 Nov 13 '21 at 10:55
  • @Patyo21 I have added further explanation hope this will help you. – Darshan P. Nov 13 '21 at 11:03
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    The further explanation made everything clear to me, now I understand it completely. Thank you very much, I have marked your solution as accepted. – Patyo21 Nov 13 '21 at 11:17
  • Why don't you mention the term "finite sum of a geometric series" ? Besides computation and proof, terminology is also an important part of the knowledge we have to convey to a student. – Jean Marie Dec 28 '21 at 09:21
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Start by: $$ S(m) = \sum_{k=0}^m N^k = 1+N+N^2+...+N^m \tag{1}\label{1} $$ Now multiply equation $\eqref{1}$ by $N$, which gives us: $$ NS(m) = N + N^2 + N^3 +... + N^m + N^{m+1}\tag{2}\label{2} $$ Now $\eqref{2} - \eqref{1}$ is equal to: $$ S(m) - NS(m) = 1 + (N-N) + (N^2-N^2) +... + (N^m - N^m) -N^{m+1} = 1-N^{m+1}\Rightarrow S(m)(1-N) = 1-N^{m+1} $$ Now if $N\ne 1\implies \boxed{S(m) = \frac{1-N^{m+1}}{1-N}}$

If $N = 1\implies \boxed{S(m)= m + 1}$, this is because $s(m)=\sum_{k=0}^m 1 = \underbrace{1 + 1 + 1 + ... + 1}_{m + 1 \text{ times}}$

aliberro
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