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Trying to formulate a proof for that sequence as practice. After reading this question's answer and lecture on this, I decided to try and practice with this sequence. My try:

Base case $n = 1$

$S(n) = 3^n $

Induction step

$S(n+1) = 3^{n+1}$

Then

$Sn + n+1 = S(n+1)$

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anchorman
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    You can use $ a^n -1 = (a-1)(1+a^2+a^3+...+a^{n-1}) $ – N.S.JOHN Sep 17 '16 at 14:45
  • Are you required to use induction? Regardless your answer is incorrect. The sum is not equal to $3^n$. This is in fact the $n^{th}$ term. Please clarify your question. – Ian Miller Sep 17 '16 at 14:57

2 Answers2

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$$\sum_{k = 1}^n\ 3^k = \frac{3}{2}\left(3^{n}-1\right)$$

Enrico M.
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$$\begin{align} \require{cancel}S & =3+9+27+\dots3^n \\ 3S & =3(3+9+27+\dots3^n) \\ & = 9+27+81+\dots3^{n+1} \\ \hline S-3 & =(3+9+27+\dots3^n)-3 \\ & = 9+27+\dots3^n \\ \hline 3S-(S-3) & = \left(\color{blue}{9+27+\dots3^n}+3^{n+1}\right)-\left(\color{blue}{9+27+\dots3^n}\right) \\ 2S+3 & = \left(\xcancel{9+27+\dots3^n}+3^{n+1}\right)-\left(\xcancel{9+27+\dots3^n}\right) \\ 2S+3 & = 3^{n+1} \\ 2S & = 3^{n+1}-3 \\ S & = \frac12\left(3^{n+1}-3\right) \end{align}$$

$$3+9+27+\dots3^n=\frac12\left(3^{n+1}-3\right)$$

Simply Beautiful Art
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