<s>[INST] Prove by induction $3+3 \cdot 5+ \cdots +3 \cdot 5^n = \frac{3(5^{n+1} -1)}{4}$ My question is:
Prove by induction that $$3+3 \cdot 5+ 3 \cdot 5^2+ \cdots +3 \cdot 5^n = \frac{3(5^{n+1} -1)}{4}$$ whenever $n$ is a nonnegative integer. 
I'm stuck at the basis step. 
If I started with $1$. I get the right hand side is $18$ which is clearly not even close. It says prove shouldn't it be always true?
 [/INST] Let $S(n)$ be the statement: $3+3\cdot{5}+\cdots+3\cdot{5^{n}}=\dfrac{3\hspace{1 mm}(5^{n+1}-1)}{4}$; $n\geq{0}$
Basis step: $S(0)$:
LHS: $3\cdot{5^{(0)}}=3$
RHS: $\dfrac{3\hspace{1 mm}(5^{(0)+1}-1)}{4}=\dfrac{3\hspace{1 mm}(5^{1}-1)}{4}$
$\hspace{35.5 mm}=\dfrac{3(4)}{4}$
$\hspace{35.5 mm}=3$
$\hspace{52.5 mm}$LHS $=$ RHS $\hspace{1 mm}$ (verified.)
Inductive step:
Assume $S(k)$ is true, i.e. assume that $3+3\cdot{5}+\cdots+3\cdot{5^{k}}=\dfrac{3\hspace{1 mm}(5^{k+1}-1)}{4}$; $k\geq{0}$
$S(k+1)$: $\underline{3+3\cdot{5}+\cdots+3\cdot{5^{k}}}+3\cdot{5^{k+1}}$
$\hspace{12.5 mm}=\dfrac{3\hspace{1 mm}(5^{k+1}-1)}{4}+3\cdot{5^{k+1}}$
$\hspace{12.5 mm}=\dfrac{3\hspace{1 mm}(5^{k+1}-1)+12\cdot{5^{k+1}}}{4}$
$\hspace{12.5 mm}=\dfrac{3\cdot{5^{k+1}}-3+12\cdot{5^{k+1}}}{4}$
$\hspace{12.5 mm}=\dfrac{15\cdot{5^{k+1}}-3}{4}$
$\hspace{12.5 mm}=\dfrac{3\cdot{5\cdot{5^{k+1}}}-3}{4}$
$\hspace{12.5 mm}=\dfrac{3\cdot{5^{k+2}}-3}{4}$
$\hspace{12.5 mm}=\dfrac{3\hspace{1 mm}(5^{k+2}-1)}{4}$
So, $S(k+1)$ is true whenever $S(k)$ is true.
Therefore, $3+3\cdot{5}+\cdots+3\cdot{5^{n}}=\dfrac{3\hspace{1 mm}(5^{n+1}-1)}{4}$; $n\geq{0}$.
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