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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)$

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., $n$.