Question

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

Answer 1

Step $\left(1\right):$ Assume given statement
Let the given statement be $P\left(n\right)$ i.e.
$P\left(n\right):(1+\frac{3}{1})(1+\frac{5}{4})(1+\frac{7}{9})\cdots (1+\frac{(2n+1)}{n^2})=(n+1{)}^2$

Step $\left(2\right):$ Checking statement $P\left(n\right)$ for $n=1$
Put $n=1$ in $P\left(n\right)$, we get
$P\left(1\right):(1+\frac{3}{1})=(1+1{)}^2$
$\Rightarrow 4=4$
Thus $P\left(n\right)$ is true for $n=1$

Step $\left(3\right):$ $P\left(n\right)$ for $n=K$
Put $n=K$ in $P\left(n\right)$ and assume this is true for some natural number $K$ i.e.,
$P\left(K\right):(1+\frac{3}{1})(1+\frac{5}{4})(1+\frac{7}{9})\cdots (1+\frac{(2K+1)}{K^2})=(K+1{)}^2\cdots (1)$

Step $\left(4\right):$ Checking statement $P\left(n\right)$ for $n=K+1$
Now we shall prove that $P(K+1)$ is true whenever $P\left(K\right)$ is true.
Now we have
$(1+\frac{3}{1})(1+\frac{5}{4})(1+\frac{7}{9})\cdots (1+\frac{(2K+1)}{K^2})(1+\frac{(2K+3)}{(K+1{)}^2})$
$=(K+1{)}^2[1+\frac{(2K+3)}{(K+1{)}^2}]$ (using$\left(1\right)$)
$=(K+1{)}^2\left[\frac{(K+1{)}^2+(2K+3)}{(K+1{)}^2}\right]$
$=K^2+2K+1+2K+3$
$=K^2+4K+4$
$=(K+2{)}^2$
We can write it as $\left[\right(K+1)+1{]}^2$
Thus, $P(K+1)$ is true whether $P\left(K\right)$ is true.
Final Answer:
Therefore, by the principle of mathematical induction, statement $P\left(n\right)$ is true for all $n\in N$ .