Prove the following by using the principle of mathematical induction for all $n\in\mathbb{N}$ :Q: 13    $\left ( 1+\frac{3}{1} \right )\left ( 1+\frac{5}{4} \right )\left ( 1+\frac{7}{9} \right )..\left ( 1+\frac{(2n+1)}{n^2} \right )=(n+1)^2$

G Gautam harsolia

Let the given statement be p(n) i.e.
$p(n):\left ( 1+\frac{3}{1} \right )\left ( 1+\frac{5}{4} \right )\left ( 1+\frac{7}{9} \right )..\left ( 1+\frac{(2n+1)}{n^2} \right )=(n+1)^2$
For n = 1  we have
$p(1):\left ( 1+\frac{3}{1} \right )= 4=(1+1)^2=2^2=4$   ,   which is true

For  n = k  we have

$p(k):\left ( 1+\frac{3}{1} \right )\left ( 1+\frac{5}{4} \right )\left ( 1+\frac{7}{9} \right )..\left ( 1+\frac{(2k+1)}{k^2} \right )=(k+1)^2 \ \ \ \ \ \ \ \ -(i)$   ,        Let's assume that this statement is true

Now,
For  n = k + 1  we have
$p(k+1):\left ( 1+\frac{3}{1} \right )\left ( 1+\frac{5}{4} \right )\left ( 1+\frac{7}{9} \right )..\left ( 1+\frac{(2(k+1)+1)}{(k+1)^2} \right )$                                                                                                                                                                                                                 $=\left ( 1+\frac{3}{1} \right )\left ( 1+\frac{5}{4} \right )\left ( 1+\frac{7}{9} \right )..\left ( 1+\frac{2k+1}{k^2} \right )\left ( 1+\frac{(2(k+1)+1)}{(k+1)^2} \right )$

$=(k+1)^2\left ( 1+\frac{(2(k+1)+1)}{(k+1)^2} \right ) \ \ \ \ \ \ \ \ \ \ (using \ (i))$
$=(k+1)^2\left ( \frac{{}(k+1)^2+(2(k+1)+1)}{(k+1)^2} \right )$
$=(k^2+1+2k+2k+2+1)$
$=(k^2+4k+4)$
$=(k+2)^2$
$=(k+1+1)^2$

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 n

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