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

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Let the given statement be $P(n)$, i.e.,
$P(n) : \bigg( 1+\large\frac{1}{1} \bigg)$$\bigg( 1+\large\frac{1}{2} \bigg)$$\bigg( 1+\large\frac{1}{3} \bigg)$$...\bigg( 1+\large\frac{1}{n} \bigg)$$=(n+1)$
For $n = 1$, we have
$P(1) : \bigg( 1+ \large\frac{1}{1} \bigg)$$=2=(1+1)$, which is true.
Let $P(k)$ be true for some positive integer $k$, i.e.,
$P(k) : \bigg( 1+\large\frac{1}{1} \bigg)$$\bigg( 1+\large\frac{1}{2} \bigg)$$\bigg( 1+\large\frac{1}{3} \bigg)$$...\bigg( 1+\large\frac{1}{k} \bigg)$$=(k+1)$-----------(i)
We shall now prove that $P(k+1)$ is true.
Consider
$\bigg[ \bigg( 1+\large\frac{1}{1} \bigg)$$\bigg( 1+\large\frac{1}{2} \bigg)$$\bigg( 1+\large\frac{1}{3} \bigg)$$...\bigg( 1+\large\frac{1}{k} \bigg)\bigg]$$ \bigg(1+ \large\frac{1}{k+1} \bigg)$
$ = (k+1) \bigg(1+ \large\frac{1}{k+1} \bigg)\qquad $ [ Using (i) ]
$ = (k+1) \bigg( \large\frac{(k+1)+1}{(k+1)} \bigg)$
$ = (k+1)+1$
Thus, $P(k+1)$ is true whenever $P(k)$ is true.
Hence, by the principle of mathematical induction, statement $P(n)$ is true for all natural numbers i.e., $n$.
answered Apr 30, 2014 by thanvigandhi_1
 

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