Prove the following by using the principle of mathematical induction for all n ∈ N:

Let the given statement be P(n), i.e.,

$\mathrm{P}(n): \frac{1}{1.4}+\frac{1}{4.7}+\frac{1}{7.10}+\ldots+\frac{1}{(3 n-2)(3 n+1)}=\frac{n}{(3 n+1)}$

For $n=1$, we have

$P(1)=\frac{1}{1.4}=\frac{1}{3.1+1}=\frac{1}{4}=\frac{1}{1.4}$, which is true.

Let P(k) be true for some positive integer k, i.e.,

$\mathrm{P}(k)=\frac{1}{1.4}+\frac{1}{4.7}+\frac{1}{7.10}+\ldots+\frac{1}{(3 k-2)(3 k+1)}=\frac{k}{3 k+1}$  (1)

We shall now prove that P(k + 1) is true.

Consider

$\left\{\frac{1}{1.4}+\frac{1}{4.7}+\frac{1}{7.10}+\ldots+\frac{1}{(3 k-2)(3 k+1)}\right\}+\frac{1}{\{3(k+1)-2\}\{3(k+1)+1\}}$

$=\frac{k}{3 k+1}+\frac{1}{(3 k+1)(3 k+4)}$  [Using (1)]

$=\frac{1}{(3 k+1)}\left\{k+\frac{1}{(3 k+4)}\right\}$

$=\frac{1}{(3 k+1)}\left\{\frac{k(3 k+4)+1}{(3 k+4)}\right\}$

$=\frac{1}{(3 k+1)}\left\{\frac{3 k^{2}+4 k+1}{(3 k+4)}\right\}$

$=\frac{1}{(3 k+1)}\left\{\frac{3 k^{2}+3 k+k+1}{(3 k+4)}\right\}$

$=\frac{(3 k+1)(k+1)}{(3 k+1)(3 k+4)}$

$=\frac{(k+1)}{3(k+1)+1}$

Thus, P(k + 1) is true whenever P(k) is true.

Hence, by the principle of mathematical induction, statement P(n) is true for all natural numbers i.e., n.