To Prove:
\(\frac1{2\times5}\) + \(\frac1{(5\times8)}\) +....+ \(\frac{1}{(3n-1)\times(3n+2)}\) = \(\frac{n}{(6n+4)}\)
For n = 1
LHS = \(\frac1{2\times5}\) = \(\frac1{10}\)
RHS = \(\frac{1\times1}{(6+4)}\) = \(\frac1{10}\)
Hence, LHS = RHS
P(n) is true for n = 1
Assume P(k) is true
\(\frac1{2\times5}\) + \(\frac1{(5\times8)}\) +....+ \(\frac{1}{(3k-1)\times(3k+2)}\) = \(\frac{k}{(6k+4)}\)......(1)
We will prove that P(k + 1) is true
RHS = \(\frac{k+1}{(6(k+1)+4)}\) = \(\frac{k+1}{(6k+10)}\)
LHS = \(\frac1{2\times5}\) + \(\frac1{(5\times8)}\) +....+ \(\frac{1}{(3k-1)\times(3k+2)}\) + \(\frac{1}{(3(k+1)-1)\times(3(k+1)+2)}\)
[Writing the Last second term]
(Taking LCM and simplifying)
= \(\frac{k+1}{(6k+10)}\)
= RHS
Therefore, \(\frac1{2\times5}\) + \(\frac1{(5\times8)}\) +....+ \(\frac{1}{(3k-1)\times(3k+2)}\) + \(\frac{1}{(3(k+1)-1)\times(3(k+1)+2)}\) = \(\frac{k+1}{(6k+10)}\)
LHS = RHS
Therefore, P (k + 1) is true whenever P(k) is true.
By the principle of mathematical induction, P(n) is true for
Where n is a natural number
Put k = n - 1
\(\frac1{2\times5}\) + \(\frac1{(5\times8)}\) +....+ \(\frac{1}{(3n-1)\times(3n+2)}\) = \(\frac{n}{(6n+4)}\)
Hence proved
