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

For a positive integer $r$, $1+2+\ldots+r=\dfrac{r(r+1)}{2}$, so the $r$-th term of the given sum is $\dfrac{1}{1+2+\ldots+r}=\dfrac{2}{r(r+1)}$.

Let $P(n)$ be the statement

Base case: For $n=1$, LHS $=1$ (the single term), and

So $P(1)$ is true.

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

We must show the sum up to the $(k+1)$-th term equals $\dfrac{2(k+1)}{k+2}$.

Adding the next term $\dfrac{2}{(k+1)(k+2)}$ to both sides of the hypothesis:

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

First rewrite the general term using $1+2+\ldots+r=\frac{r(r+1)}{2}$, turning it into $\frac{2}{r(r+1)}$. Then apply the standard add-the-next-term induction: combine $\frac{2k}{k+1}$ with $\frac{2}{(k+1)(k+2)}$ over a common denominator and simplify the resulting perfect square in the numerator.

• Forgetting to convert $1+2+\ldots+r$ into $\frac{r(r+1)}{2}$ first, and instead trying to induct directly on the un-simplified denominator.
• Losing track of which term is "next": the $(k+1)$-th term is $\dfrac{2}{(k+1)(k+2)}$, not $\dfrac{2}{(k+1)(k)}$.
• Arithmetic error combining fractions over the common denominator $(k+1)(k+2)$.