Question

Prove the following by using the principle of mathematical induction for all $n\in N:\frac{1}{1.4}+\frac{1}{4.7}+\frac{1}{7.10}+.....+\frac{1}{(3n-2)(3n+1)}=\frac{n}{(3n+1)}$

Answer 1

Let the given statement be $P\left(n\right)$ i.e.,
$P\left(n\right):\frac{1}{1.4}+\frac{1}{4.7}+\frac{1}{7.10}+.....+\frac{1}{(3n-2)(3n+1)}=\frac{n}{(3n+1)}$
For $n=1$, we have
$P\left(1\right)=\frac{1}{1.4}=\frac{1}{3.1+1}=\frac{1}{4}=\frac{1}{1.4}$, which is true.
Let $P\left(k\right)$ be true for some $k\in N$, i.e.,
$P\left(k\right)=\frac{1}{1.4}+\frac{1}{4.7}+\frac{1}{7.10}+.....+\frac{1}{(3k-2)(3k+1)}=\frac{k}{(3k+1)}......\left(i\right)$
We shall now prove that $P(k+1)$ is true.
Consider
$\frac{1}{1.4}+\frac{1}{4.7}+\frac{1}{7.10}+.....+\frac{1}{(3k-2)(3k+1)}+\frac{1}{\left\{3\right(k+1)-2\}\left\{3\right(k+1)+1\}}$ [Using $\left(i\right)$]
$=\frac{k}{3k+1}+\frac{1}{(3k+1)(3k+4)}$
$=\frac{1}{(3k+1)}\{k+\frac{1}{(3k+4)}\}$
$=\frac{1}{(3k+1)}\left\{\frac{k(3k+4)+1}{(3k+4)}\right\}$
$=\frac{1}{(3k+1)}\left\{\frac{3k^2+4k+1}{(3k+4)}\right\}$
$=\frac{(3k+1)(k+1)}{(3k+1)(3k+4)}$
$=\frac{(k+1)}{3(k+1)+1}$
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., $n$.