Question

Prove the following by using the principle of mathematical induction for all $n\in N:\frac{1}{2.5}+\frac{1}{5.8}+\frac{1}{8.11}+......+\frac{1}{(3n-1)(3n+2)}=\frac{n}{(6n+4)}$

Answer 1

Let the given statement be $P\left(n\right)$ i.e
$P\left(n\right):\frac{1}{2.5}+\frac{1}{5.8}+\frac{1}{8.11}+......+\frac{1}{(3n-1)(3n+2)}=\frac{n}{(6n+4)}$
For $n=1$, we have
$P\left(1\right)=\frac{1}{2.5}=\frac{1}{10}=\frac{1}{6.1+4}+\frac{1}{10}$, which is true.
Let $P\left(k\right)$ be true for some $k\in N$ i.e.,
$\frac{1}{2.5}+\frac{1}{5.8}+\frac{1}{8.11}+......+\frac{1}{(3k-1)(3k+2)}=\frac{k}{6k+4}.......\left(i\right)$
We shall now prove that $P(k+1)$ is true.
Consider
$\frac{1}{2.5}+\frac{1}{5.8}+\frac{1}{8.11}+.......+\frac{1}{(3k-1)(3k+2)}+\frac{1}{\left\{3\right(k+1)-1\}\left\{3\right(k+1)+2\}}$
$=\frac{k}{6k+4}+\frac{1}{(3k+3-1)(3k+3+2}$
$=\frac{k}{6k+4}+\frac{1}{(3k+2)(3k+5)}$
$=\frac{1}{(3k+2)}(\frac{k}{2}+\frac{1}{3k+5})$
$=\frac{1}{(3k+2)}\left(\frac{k(3k+5)+2}{2(3k+5)}\right)$
$=\frac{1}{(3k+2)}\left(\frac{(3k+2)(k+1)}{2(3k+5)}\right)$
$=\frac{(k+1)}{6k+10}$
$=\frac{(k+1)}{6(k+1)+4}$
Thus $P(k+1)$ is true whenever $P\left(k\right)$ is true.
Hence, by the principle of mathematical induction, statement $P\left(n\right)$ is true for all natural numbers i.e., $n$