Prove by induction that $1+3+3^2+3^3+...+3^n=\frac{1}{2}(3^{n+1}-1)$

Question

I have $\frac{1}{2}(3^{k+1}-1)+3^{k+1}=\frac{1}{2}3^{k+2}-1$

Not sure where to go from here.

First post on Mathematics, please correct me if I did anything wrong.

Answer 1

There is a general formula for $S_n=1+q+q^2+\cdots+q^n$ :

if $q=1$, then $S_n=n+1$ and if $q\neq 1$ then :

$$S_n=\frac{1-q^{n+1}}{1-q}\tag{*}$$

You can prove it either by induction or by computing the product $(1-q)S_n$ : observe how terms cancel.

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We suppose $q\neq1$.

By induction :

Clearly :

$$S_0=1=\frac{1-q}{1-q}$$

so the formula (*) is true for $n=0$.

Now suppose it true for some $n\in\mathbb{N}$; we have :

$$S_{n+1}=S_n+q^{n+1}=\frac{1-q^{n+1}}{1-q}+q^{n+1}=\frac{1-q^{n+1}+q^{n+1}(1-q)}{1-q}$$

and after simplification :

$$S_{n+1}=\frac{1-q^{n+2}}{1-q}$$

as desired. So the formula (*) is true for all $n\in\mathbb{N}$.

By direct computation :

Expanding the product $(1-q)S_n$ leads to a "telescopic summation" :

$$(1-q)S_n=S_n-qS_n=\sum_{k=0}^nq^k-\sum_{k=0}^nq^{k+1}=1+\sum_{k=1}^nq^k-\left(q^{n+1}+\sum_{k=0}^{n-1}q^{k+1}\right)$$

and after simplification :

$$(1-q)S_n=1-q^{n+1}$$