Prove by mathematical induction
1 + 3 + 3^2 + ... + 3^(n-1) = [3^(n-1)] / 2
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i really need helpp, i really don't get it after the second step.
Hi phillychum,
it should read $\displaystyle 1+3+3^2+....+3^{n-1}=\frac{3^n-1}{2}$
If this is true, then
$\displaystyle 1+3+3^2+......+3^{n-1}+3^n$ should $\displaystyle =\frac{3^{n+1}-1}{2}$
We can prove if this is the case by adding $\displaystyle 3^n$ to both sides of the original equality.
$\displaystyle 1+3+3^2+....+3^{n-1}+3^n=\frac{3^n-1}{2}+3^n\ ?$
$\displaystyle \frac{3^n-1}{2}+3^n=\frac{3^n-1}{2}+\frac{(2)3^n}{2}=\frac{(3)3^n-1}{2}=\frac{3^{n+1}-1}{2}$
This means that if $\displaystyle 1+3+3^2+...+3^{n-1}=\frac{3^n-1}{2}$
then $\displaystyle 1+3+3^2+...+3^n=\frac{3^{n+1}-1}{2}$
Hence the inductive step is proven.
Now it only needs to be proven for term number 1.