Question

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

Answer 1

Let the given statement be $P\left(n\right)$, i.e.,
$P\left(n\right):(1+\frac{3}{1})(1+\frac{5}{4})(1+\frac{7}{9})......(1+\frac{(2n+1)}{n^2})={(n+1)}^2$
For $n=1$, we have
$P\left(1\right):(1+\frac{3}{1})=4={(1+1)}^2=2^2=4$, which is true.
Let $P\left(k\right)$ be true for some $k\in N$, i.e
$(1+\frac{3}{1})(1+\frac{5}{4})(1+\frac{7}{9})......(1+\frac{(2k+1)}{k^2})={(k+1)}^2..............\left(1\right)$
We shall now prove that $P(k+1)$ is true.
Consider
$(1+\frac{3}{1})(1+\frac{5}{4})(1+\frac{7}{9})......(1+\frac{(2k+1)}{k^2})(1+\frac{\left\{2\right(k+1)+1\}}{{(k+1)}^2})$
$={(k+1)}^2[1+\frac{\left\{2\right(k+1)+1\}}{{(k+1)}^2}]$ [Using $\left(i\right)$]
$={(k+1)}^2\left[\frac{{(k+1)}^2+2(k+1)+1}{{(k+1)}^2}\right]$
$={(k+1)}^2+2(k+1)+1$
$={\left\{\right(k+1)+1\}}^2$
Thus, $P(k+1)$ is true whenever $P\left(k\right)$ is true.
Hence by the principle of mathematical induction, statement $P\left(n\right)$ is true for all natural numbers i.e., $n$.