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

To Prove:

$\frac{1}{2 \times 5}+\frac{1}{(5 \times 8)}+\ldots \ldots+\frac{1}{(3 n-1) \times(3 n+2)}=\frac{n}{(6 n+4)}$

For n = 1

$\mathrm{LHS}=\frac{1}{2 \times 5}=\frac{1}{10}$

$\mathrm{RHS}=\frac{1 \times 1}{(6+4)}=\frac{1}{10}$

Hence, LHS = RHS

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

Assume $P(k)$ is true

$\frac{1}{2 \times 5}+\frac{1}{(5 \times 8)}+\ldots \ldots+\frac{1}{(3 k-1) \times(3 k+2)}=\frac{k}{(6 k+4)} \ldots \ldots(1)$

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

$\mathrm{RHS}=\frac{k+1}{(6(k+1)+4)}=\frac{k+1}{(6 k+10)}$

$\mathrm{LHS}=\frac{1}{2 \times 5}+\frac{1}{(5 \times 8)}+\ldots \ldots+\frac{1}{(3 k-1) \times(3 k+2)}+\frac{1}{(3(k+1)-1) \times(3(k+1)+2)}$ 

[Writing the Last second term]

$=\frac{1}{2 \times 5}+\frac{1}{(5 \times 8)}+\ldots \ldots+\frac{1}{(3 k-1) \times(3 k+2)}+\frac{1}{(3(k+1)-1) \times(3(k+1)+2)}$

$=\frac{k}{(6 k+4)}+\frac{1}{(3(k+1)-1) \times(3(k+1)+2)}$ [Using 1]

$=\frac{k}{(6 k+4)}+\frac{1}{(3 k+2) \times(3 k+5)}$

$=\frac{k}{(6 k+4)}+\frac{1}{(3 k+2) \times(3 k+5)}$

$=\frac{1}{(3 k+2)} \times\left[\frac{(3 k+2) \times(k+1)}{2 \times(3 k+5)}\right]$ (Taking LCM and simplifying)

$=\frac{k+1}{(6 k+10)}$

= RHS

Therefore, $\frac{1}{2 \times 5}+\frac{1}{(5 \times 8)}+\ldots \ldots+\frac{1}{(3 k-1) \times(3 k+2)}+\frac{1}{(3(k+1)-1) \times(3(k+1)+2)}=$

$\frac{k+1}{(6 k+10)}$

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

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

By the principle of mathematical induction, $\mathrm{P}(\mathrm{n})$ is true for

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

Put $k=n-1$

$\frac{1}{2 \times 5}+\frac{1}{(5 \times 8)}+\ldots \ldots \times+\frac{1}{(3 n-1) \times(3 n+2)}=\frac{n}{(6 n+4)}$

Hence proved.