Q.For all $n \in N$, let $P(n) : 1 + 3 + 3^2 + \ldots + 3^{n-1} = \dfrac{(3^n - 1)}{2}$

Verify the base case directly, then show P(k) true forces P(k+1) true.

$P(n): \dfrac12+\dfrac14+\dfrac18+\cdots+\dfrac1{2^n} = 1-\dfrac1{2^n}$

1. Base case, $n=1$: LHS $=\dfrac12$. RHS $=1-\dfrac1{2^1}=1-\dfrac12=\dfrac12$. LHS = RHS, so $P(1)$ is true.
2. Inductive step: Assume $P(k)$ is true for some $k\in N$: $\dfrac12+\dfrac14+\cdots+\dfrac1{2^k} = 1-\dfrac1{2^k}\quad\text{(inductive hypothesis)}$ Add $\dfrac1{2^{k+1}}$ to both sides: $\dfrac12+\dfrac14+\cdots+\dfrac1{2^k}+\dfrac1{2^{k+1}} = \left(1-\dfrac1{2^k}\right)+\dfrac1{2^{k+1}}$ Simplify the RHS: $1-\dfrac{2}{2^{k+1}}+\dfrac1{2^{k+1}} = 1-\dfrac1{2^{k+1}}$ …

Verify the base case, then assume $P(k)$ holds and add the next term $3^k$ to derive $P(k+1)$.

Statement

$P(n): 1+3+3^2+\cdots+3^{n-1} = \frac{3^n-1}{2}$

1. Base case: P(1) LHS of $P(1)$ is just the first term: $1$ (i.e. $3^{1-1}=3^0=1$). RHS $=\dfrac{3^1-1}{2} = \dfrac{2}{2}=1$. Since LHS $=$ RHS $=1$, $P(1)$ is true.
2. Inductive step Assume $P(k)$ is true: $1+3+3^2+\cdots+3^{k-1} = \frac{3^k-1}{2}$ Add the next term $3^k$ to both sides: $1+3+3^2+\cdots+3^{k-1}+3^k = \frac{3^k-1}{2}+3^k$ $= \frac{3^k-1+2\cdot 3^k}{2} = \frac{3\cdot 3^k - 1}{2} = \frac{3^{k+1}-1}{2}$ …