Q.Prove the following by using the principle of mathematical induction for all $n \in N$: $1 + 3 + 3^2 + \ldots + 3^{n-1} = \dfrac{3^n - 1}{2}$

Let $P(n)$ be the statement

Base case: For $n=1$, LHS $=3^0=1$ (single term), and

So $P(1)$ is true.

Inductive step: Assume $P(k)$ is true for some $k\ge1$:

We must show $1+3+\ldots+3^{k-1}+3^k=\dfrac{3^{k+1}-1}{2}$.

Adding $3^k$ (the next term) to both sides of the hypothesis:

This is exactly $P(k+1)$.

Geometric-series induction. Add the next term $3^k$ to the assumed sum $\frac{3^k-1}{2}$, combine over a common denominator, and simplify $3^k-1+2\cdot3^k=3\cdot3^k-1=3^{k+1}-1$.

• Confusing the last term of the sum up to $n=k$: it is $3^{k-1}$, so the next term to add is $3^k$, not $3^{k+1}$.
• Base-case error: taking LHS at $n=1$ to be "$1+3$" instead of just the single term $3^0=1$.
• Arithmetic slip combining $3^k-1+2\cdot3^k$ instead of correctly factoring to $3\cdot3^k-1$.