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For all $ n \geq 1$, prove that \[\] $ \large\frac{1}{1.2}$$+\large\frac{1}{2.3}$$+\large\frac{1}{3.4}$$+...+\large\frac{1}{n(n+1)}$$=\large\frac{n}{(n+1)}$.

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We can write
$P(n) : \large\frac{1}{1.2}$$+\large\frac{1}{2.3}$$+\large\frac{1}{3.4}$$+...+\large\frac{1}{n(n+1)}$$=\large\frac{n}{(n+1)}$
We note that $P(1): \large\frac{1}{1.2}$$=\large\frac{1}{2}$$ = \large\frac{1}{1+1}$, which is true. Thus, $P(n)$ is true for $n=1$.
Assume that $P(k)$ is true for some natural number $k$,
i.e., $ \large\frac{1}{1.2}$$+\large\frac{1}{2.3}$$+\large\frac{1}{3.4}$$+...+\large\frac{1}{k(k+1)}$$=\large\frac{k}{(k+1)}$------------(1)
We need to prove that $P(k+1)$ is true whenever $P(k)$ is true. We have
$ \large\frac{1}{1.2}$$+\large\frac{1}{2.3}$$+\large\frac{1}{3.4}$$+...+\large\frac{1}{k(k+1)}$$+\large\frac{1}{(k+1)(k+2)}$
$ \bigg[ \large\frac{1}{1.2}$$+\large\frac{1}{2.3}$$+\large\frac{1}{3.4}$$+...+\large\frac{1}{k(k+1)}\bigg]$$+\large\frac{1}{(k+1)(k+2)}$
$ = \large\frac{k}{k+1}$$+ \large\frac{1}{(k+1)(k+2)}\qquad$ [ Using (1) ]
$ = \large\frac{k(k+2)+1}{(k+1)(k+2)}$$=\large\frac{(k^2+2k+1)}{(k+1)(k+2)}$$ = \large\frac{(k+1)^2}{(k+1)(k+2)}$$ = \large\frac{k+1}{k+2}$$=\large\frac{k+1}{(k+1)+1}$
Thus, $P(k+1)$ is true whenever $P(k)$ is true. Hence, by the principle of mathematical induction, $P(n)$ is true for all natural numbers.
answered May 2, 2014 by thanvigandhi_1

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