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Prove the following by using the principle of mathematical induction for all $n \in N$ \[\] $1+ \large\frac{1}{(1+2)}$$+ \large\frac{1}{(1+2+3)}$$+...+ \large\frac{1}{(1+2+3...n)}$$= \large\frac{2n}{(n+1)}$

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Let the given statement be $P(n)$, i.e.,
$P(n) : 1+ \large\frac{1}{1+2}$$+\large\frac{1}{1+2+3}$$ +...+ \large\frac{1}{1+2+3+...n} $$ = \large\frac{2n}{n+1}$
For $n=1$, we have
$P(1) : 1 = \large\frac{2.1}{1+1}$$ = \large\frac{2}{2}$$ =1$, which is true.
Let $P(k)$ be true for some positive integer $k$, i.e.,
$= 1+ \large\frac{1}{1+2}$$+\large\frac{1}{1+2+3}$$ +...+ \large\frac{1}{1+2+3+...+k} $$ = \large\frac{2k}{k+1}$---------(i)
We shall now prove that $P(k+1)$ is true.
Consider
$= 1+ \large\frac{1}{1+2}$$+\large\frac{1}{1+2+3}$$ +...+ \large\frac{1}{1+2+3+...+k} $$+\large\frac{1}{1+2+3+...+k+(k+1)} $
$=\bigg( 1+ \large\frac{1}{1+2}$$+\large\frac{1}{1+2+3}$$ +...+ \large\frac{1}{1+2+3+...+k}\bigg) $$+\large\frac{1}{1+2+3+...+k+(k+1)} $
$ = \large\frac{2k}{k+1} $$+ \large\frac{1}{1+2+3+...+k+(k+1)} \qquad $ [ Using (i) ]
$ = \large\frac{2k}{k+1}$$+ \large\frac{1}{ \bigg( \Large\frac{(k+1)(k+1+1)}{2} \bigg)}$$ \qquad \bigg[1+2+3+..+n = \large\frac{n(n+1)}{2} \bigg]$
$ = \large\frac{2k}{k+1} $$ + \large\frac{2}{(k+1)(k+2)}$
$ \large\frac{2}{(k+1)}$$ \bigg( k+ \large\frac{1}{k+2}\bigg)$
answered Apr 29, 2014 by thanvigandhi_1
 

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