problem proof of induction $2\cdot3^0+2\cdot3^1+2\cdot3^2+...+2\cdot3^{n-1}=3^n-1$

Question

I need to prove that $2\cdot3^0+2\cdot3^1+2\cdot3^2+...+2\cdot3^{n-1}=3^n-1,\, n=1,2,3,4...$

I know when n = 1 the left side $=2\cdot3^{1-1}=2$. The right side is $2^n - 1=2^1 - 1 = 1$. Thus the statement is true for $n = 1$.

True for $n = k$, that is $2\cdot3^0+2\cdot3^1+2\cdot3^2+...+2\cdot3^{k-1}=3^k-1$

I have problem to prove that the statement is true for $n = k + 1$. I don't know how I should continue...

Answer 1

So we assume that $$2\cdot3^0+2\cdot3^1+2\cdot3^2+...+2\cdot3^{k-1}=3^k-1$$ Now add $2 \cdot 3^{k}$ to both sides: $$ 2\cdot3^0+2\cdot3^1+2\cdot3^2+...+2\cdot3^{k-1}+2\cdot 3^{k}=2\cdot 3^k +3^k-1 = 3 \cdot 3^{k}-1 = 3^{k+1}-1, $$ which is of the same form as the $n=k$ case with $k$ replaced by $k+1$, and hence the statement is true for $n=k+1$.

Answer 2

$2 \times3^0+2 \times 3^1+...+2 \times 3^{n-1}+2 \times 3^n=3^n-1+2 \times 3^n=3^{n+1}-1$

Hope this helps.