Using the principle of mathematical induction, prove each of the following

To Prove:

$\frac{1}{1 \times 4}+\frac{1}{(4 \times 7)}+\ldots \ldots+\frac{1}{(3 n-2) \times(3 n+1)}=\frac{n}{(3 n+1)}$

Let us prove this question by principle of mathematical induction (PMI)

Let $P(n): \frac{1}{1 \times 4}+\frac{1}{(4 \times 7)}+\ldots \ldots+\frac{1}{(3 n-2) \times(3 n+1)}=\frac{n}{(3 n+1)}$

For n = 1

$\mathrm{LHS}=\frac{1}{1 \times 4}=\frac{1}{4}$

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

Hence, $\mathrm{LHS}=\mathrm{RHS}$

$P(n)$ is true for $n=1$

Assume $P(k)$ is true

$=\frac{1}{1 \times 4}+\frac{1}{(4 \times 7)}+\ldots \ldots+\frac{1}{(3 k-2) \times(3 k+1)}=\frac{k}{(3 k+1)} \ldots \ldots(1)$

We will prove that P(k + 1) is true

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

$\mathrm{LHS}=\frac{1}{1 \times 4}+\frac{1}{(4 \times 7)}+\ldots \ldots+\frac{1}{(3(k+1)-2) \times(3(k+1)+1)}$

$=\frac{1}{1 \times 4}+\frac{1}{(4 \times 7)}+\ldots \ldots+\frac{1}{(3 k-2) \times(3 k+1)}+\frac{1}{(3 k+1) \times(3 k+4)}$

[Writing the second last term]

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

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

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

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

(Splitting the numerator and cancelling the common factor)

$=\mathrm{RHS}$

$\mathrm{LHS}=\mathrm{RHS}$

Therefore, $P(k+1)$ is true whenever $P(k)$ is true.

By the principle of mathematical induction, $P(n)$ is true for

Where $\mathrm{n}$ is a natural number

Hence proved.