Let T(n) be the statement:
\(\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{n(n+1)} = \frac{n}{n+1}\)
Basic Step: For n = 1, \(\frac{1}{1.2} = \frac{1}{1.(1+1)}\)
⇒ T (1) is true.
Induction Step: Assume T (k) is true, i.e.,
\(\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{k(k+1)} = \frac{k}{k+1}\)
To obtain T(k + 1), add the (k + 1)th term, i.e., \(\frac{1}{(k+1)(k+2)}\) to both sides of T(k). Then,
\(\frac{1}{1.2}+\frac{1}{2.3}+...\frac{1}{k(k+1)}+\frac{1}{(k+1)(k+2)}=\frac{k}{k+1}+\frac{1}{(k+1)(k+2)}\)
\(\frac{k(k+2)+1}{(k+1)(k+2)} = \frac{k^2+2k+1}{(k+1)(k+2)} = \frac{(k+1)^2}{(k+1)(k+2)} =\frac{(k+1)}{(k+2)}\)
Thus T (k + 1) is true with the assumption that T(k) is true.
Hence the statement T(n) holds for all positive integral values of n.
