Can someone verify my proof by induction for the following:
$2+6+18+…+2 \cdot 3^{n-1} = 3^n – 1, n\geq 1$.
Basis $n = 1$:
\begin{align*}
2+6+18+…+2 \cdot 3^{n-1} = 3^n – 1 &= \sum\limits_{i=1}^{n} 2 \cdot 3^{n-1} = 3^n -1 \\
2+6+18+…+2 \cdot 3^{1-1} &= 3^1 – 1 \\
\sum\limits_{i=1}^{n} 2 \cdot 3^{1-1} &= 3^1 -1 \\
2 \cdot 1 = 3-1 \\
2 = 2 \checkmark \\
\end{align*}
Inductive Hypothesis: Assume $P(x)$ is true, i.e.,
\begin{align*}
\sum\limits_{i=1}^{n} 2 \cdot 3^{i-1} &= 3^k -1 \\
\end{align*}
Inductive Step: Show $P(k+1)$ is true, i.e.,
\begin{align*}
\sum\limits_{i=1}^{k+1} 2 \cdot 3^{i-1} &= 3^{k+1} -1 \\
\sum\limits_{i=1}^{k+1} 2 \cdot 3^{i} &= 3^{k+1} -1 \\
\end{align*}
Best Answer
What I would write:
Proposition to be proven $P(n)$ is that $2 + 6 + 18 + ... + 2\cdot3^{n-1} = 3^n-1, \forall n \in \mathbb {Z^+}$, which may also be written $\displaystyle \sum_{i=1}^n 2\cdot 3^{i-1} = 3^n-1, \forall n \in \mathbb{Z^+}$.
Basis step: Consider $P(1)$.
For $n=1, \mathrm{LHS\ (left \ hand \ side)} = 2\cdot 3^{1-1} = 2$
While $\mathrm{RHS} = 3^1 - 1 = 2$.
Hence $\mathrm{LHS} = \mathrm{RHS}$ and $P(1)$ is true.
Inductive step:
Inductive hypothesis: assume $P(k)$ holds for some $k \in \mathbb{Z^+}$. This may be written $\displaystyle \sum_{i=1}^k 2\cdot 3^{i-1} = 3^k-1$ for a particular $k \in \mathbb{Z^+}$.
Now consider $P(k+1)$:
$\mathrm{LHS}_{P(k+1)} \\= \displaystyle \sum_{i=1}^{k+1} 2\cdot 3^{i-1}\\ = 2\cdot 3^{k+1-1} + \displaystyle \sum_{i=1}^{k} 2\cdot 3^{i-1} \\= 2\cdot 3^{k} + \displaystyle \sum_{i=1}^{k} 2\cdot 3^{i-1} \\ = 2\cdot 3^{k} + 3^k - 1 \mathrm{\ (applying\ inductive\ hypothesis)}\\ = 3 \cdot 3^{k} - 1 \\= 3^{k+1} - 1 \\=\mathrm{RHS}_{P(k+1)}$
Hence $P(k) \implies P(k+1)$.
By induction, $P(n)$ is true $\forall n \in \mathbb{Z^+}$ (QED)