Question

Prove the following by using the principle of mathematical induction for all $n\in$:
$1+\frac{1}{(1+2)}+\frac{1}{(1+2+3)}+..........+\frac{1}{1+2+3+.....n)}=\frac{2n}{(n+1)}$

Answer 1

Let the given statement be $P\left(n\right)$ i.e.,
$P\left(n\right)$:
For $n=1$, we have
$P\left(1\right):1=\frac{2\times 1}{1+1}=\frac{2}{2}=1$
which is true.
Let $P\left(k\right)$ be true for some $k\in N$,
$1+\frac{1}{1+2}+......\frac{1}{1+2+3}+......+\frac{1}{1+2+3+.....+k}=\frac{2k}{k+1}.........\left(i\right)$
We shall now prove that $P(k+1)$ is true.
Consider
$1+\frac{1}{1+2}+\frac{1}{1+2+3}+.......+\frac{1}{1+2+3+....+k}+\frac{1}{1+2+3+......+k+(k+1)}$
$=(1+\frac{1}{1+2}+\frac{1}{1+2+3}+.......+\frac{1}{1+2+3+....k})+\left(\frac{1}{1+2+3+.......+k+(k+1)}\right)$
$=\frac{2k}{k+1}+\frac{1}{1+2+3+.....+k+(k+1)}$ [Using $\left(i\right)$]
$=\frac{2k}{k+1}+\frac{1}{\left(\frac{(k+1)(k+1+1)}{2}\right)}$ $(\because 1+2+3+.....+n=\frac{n(n+1)}{2})$
$=\frac{2k}{(k+1)}+\frac{2}{(k+1)(k+2)}$
$=\frac{2}{(k+1)}(k+\frac{1}{k+2})$
$=\frac{2(k^2+2k+1)}{(k+1)(k+2)}$
$=\frac{2(k+1)}{k+2}$
Hence, the given result is true for $k+1$.
So, by mathematical induction method, the given result is true for all $n\in N$