How continue with this proof?

Question

I'm trying to proof the following statement:

$7+6 \cdot (7+7^1+7^2+\dots+7^n) = 7^{n+1}$

I try the next: $n=k$

$7+6 \cdot (7+7^1+7^2+\dots+7^k) = 7^{k+1}$

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

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

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

$n=k+1$

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

Proof:

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

$\frac{7^{k+1}-7+6\cdot7^{k+1}}{6}$

Any idea?

Answer 1

You asked for a proof, so here's proof by induction.

Your statement is equivalent to:

$$6 \cdot \sum_{k=1}^n 7^k = 7^{n+1} - 7.$$

This is true for $n=1$:

$$6 \cdot 7 = 7^2 - 7.$$

Now look at the $n+1$ case:

$$6 \cdot \sum_{k=1}^{n+1} 7^k = \left[6 \cdot \sum_{k=1}^{n} 7^k\right] + 6 \cdot 7^{n+1} \\ = 7^{n+1} - 7 + 6 \cdot 7^{n+1} \\= 7 \cdot 7^{n+1} - 7 \\= 7^{n+2} - 7.$$

I invoked the inductive assumption on the second line.

Answer 2

Use the formula for the sum of a geometric sequence:

$$7+6\left(7+7^2+\ldots+7^n\right)=7+6\frac{7-7^{n+1}}{1-7}=7-7+7^{n+1}=7^{n+1}$$

Answer 3

It seems you want to demonstrate by induction that
$$ 7 + 6 \cdot \left( {7 + 7^{\,2\,} + \cdots + 7^{\,n\,} } \right) = 7^{\,n + 1\,} $$
Now, that's true for $n=1$ $$ \begin{gathered} n = 1 \hfill \\ 7 + 6 \cdot \left( 7 \right) = 7 \cdot \left( {6 + 1} \right) = 7^{\,2\,} \hfill \\ \end{gathered} $$
and we want to demonstrate that if it is valid for $n$ it is also valid for $n+1$.
So we take our identity, multiply both sides by $7$, do some massage $$ \begin{gathered} n \to n + 1 \hfill \\ 7 + 6 \cdot \left( {7 + 7^{\,2\,} + \cdots + 7^{\,n\,} } \right) = 7^{\,n + 1\,} \hfill \\ 7 \cdot \left( {7 + 6 \cdot \left( {7 + 7^{\,2\,} + \cdots + 7^{\,n\,} } \right)} \right) = 7 \cdot 7^{\,n + 1\,} = 7^{\,n + 2\,} \hfill \\ \left( {7 \cdot 7 + 6 \cdot 7 \cdot \left( {7 + 7^{\,2\,} + \cdots + 7^{\,n\,} } \right)} \right) = 7^{\,n + 2\,} \hfill \\ \left( {7 \cdot \left( {6 + 1} \right) + 6 \cdot 7 \cdot \left( {7 + 7^{\,2\,} + \cdots + 7^{\,n\,} } \right)} \right) = 7^{\,n + 2\,} \hfill \\ \left( {7 + 6 \cdot 7 + 6 \cdot \left( {7^{\,2\,} + 7^{\,3\,} + \cdots + 7^{\,n + 1\,} } \right)} \right) = 7^{\,n + 2\,} \hfill \\ \left( {7 + 6 \cdot \left( {7 + 7^{\,2\,} + 7^{\,3\,} + \cdots + 7^{\,n + 1\,} } \right)} \right) = 7^{\,n + 2\,} \hfill \\ \end{gathered} $$
and get the demonstration