Prove using simple induction

Question

Prove by induction for each integer $n$ greater or equal to $1$, we have $$7 + 7^2 + 7^3 +···+ 7^n = \frac{7^{n+1}-7}{6}.$$

Answer 1

The statement is clear at $n=1$. Assume that the statement holds for $n=k$. Then $$7+7^2+7^3+\dots+7^k=\frac{7^{k+1}-7}{6}.$$ Now $$\begin{align}7+7^2+7^3+\dots+7^\color\red{{k+1}}&=(7+7^2+7^3+\dots+7^k)+7^{k+1}\\ &=\frac{7^{k+1}-7}{6}+7^{k+1}\\ &=\frac{1(7^{k+1})-7+6(7^{k+1})}{6}\\ &=\frac{7(7^{k+1})-7}{6}\\ &=\frac{7^{(\color\red{k+1})+1}-7}{6}. \end{align}$$ Hence, the statement holds at $n=k+1$. Hope it helps.

Answer 2

First, prove the base case, that is, $n=1$: $7^1=\frac{7^2-7}{6}=7$

Then, rewrite for inductive case (i.e. for $n=k$):

$7+7^2+...+7^k=\frac{7^{(k+1)}-7}{6}$

We then form the inductive hypothesis (i.e. $n=k+1$):

$7+7^2+...+7^k+7^{k+1}=\frac{7^{(k+2)}-7}{6}$

This gives us the following:

$\frac{7^{(k+1)}-7}{6}+7^{k+1}=\frac{7^{(k+2)}-7}{6}$

Simplify:

$\frac{7^{k+1}-7+6\bullet7^{k+1}}{6}=\frac{7^{(k+2)}-7}{6}$

Factor out $7^{k+1}$:

Giving us the final answer,

$\frac{7^{(k+2)}-7}{6}=\frac{7^{(k+2)}-7}{6}$