To Prove:
\(\frac{1}{1\times4}\) + \(\frac{1}{4\times7}\) + \(\frac{1}{7.10}\)+......+ \(\frac{1}{(3n - 2)\times(3n+1)}\) = \(\frac{n}{(3n +1)}\)
Let us prove this question by principle of mathematical induction (PMI)
Let P(n): \(\frac{1}{1\times4}\) + \(\frac{1}{4\times7}\) +......+ \(\frac{1}{(3n - 2)\times(3n+1)}\) = \(\frac{n}{(3n +1)}\)
For n = 1
LHS = \(\frac{1}{1\times4}\) = \(\frac{1}{4}\)
RHS = \(\frac{1}{(3+1)}\) = \(\frac{1}{4}\)
Hence, LHS = RHS
P(n) is true for n = 1
Assume P(k) is true
= \(\frac{1}{1\times4}\) + \(\frac{1}{(4\times7)}\) +......+ \(\frac{1}{(3k - 2)\times(3k+1)}\) = \(\frac{k}{(3k+1)}\)......(1)
We will prove that P(k + 1) is true
RHS = \(\frac{k+1}{(3(k+1)+1)}\) = \(\frac{k+1}{(3k+4)}\)
[Writing the second last term]
(Splitting the numerator and cancelling the common factor)
= RHS
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
Hence proved.
