To Prove:
\(\frac{1}1\) + \(\frac{1}{(1 +2)}\) + \(\frac{1}{(1 +2+3+.......+n)}\) = \(\frac{2n}{(n+1)}\)
Let us prove this question by principle of mathematical induction (PMI)
Let P(n): \(\frac{1}1\) + \(\frac{1}{(1 +2)}\) + \(\frac{1}{(1 +2+3+.....+n)}\) = \(\frac{2n}{(n+1)}\)
For n = 1
LHS = 1
RHS = \(\frac{2 \times 1}{(1+1)}\) = 1
Hence, LHS = RHS
P(n) is true for n = 1
Assume P(k) is true
\(\frac{1}1\) + \(\frac{1}{(1 +2)}\) + \(\frac{1}{(1 +2+3+......\times+k)}\) = \(\frac{2k}{(k+1)}\)......(1)
We will prove that P(k + 1) is true
RHS = \(\frac{2(k+1)}{(k+1+1)}\) = \(\frac{2k+2}{(k+2)}\)
LHS = \(\frac{1}1\) + \(\frac{1}{(1 +2)}\) +.......+ \(\frac{1}{(1 +2+3+......+(k+1))}\)
= \(\frac{1}1\) + \(\frac{1}{(1 +2)}\) +.......+ \(\frac{1}{(1 +2+3+......+k)}\) + \(\frac{1}{(1 +2+3+......+(k+1))}\)
[Writing the last Second term]
[Taking LCM and simplifying]
= \(\frac{2(k+1)}{(k+2)}\)
= RHS
Therefore, \(\frac{1}1\) + \(\frac{1}{(1 +2)}\) +.......+ \(\frac{1}{(1 +2+3+......\times +(k+1))}\) = \(\frac{2k+2}{k+2}\)
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
\(\frac{1}1\) + \(\frac{1}{(1 +2)}\) +.......+ \(\frac{1}{(1 +2+3+......\times +n)}\) = \(\frac{2n}{n+1}\)
Hence proved
