First slide

Introduction to P.M.I

Question

The value of $(1 + \frac{3}{1}) (1 + \frac{5}{4}) (1 + \frac{7}{9}) \dots (1 + \frac{2 n + 1}{n^{2}})$ is

Moderate

A

(n+1)
B

(n+1)²
C

2(n+1)²
D

None of these

Solution

Let the given statement be P(n). i.e
$P (n) : (1 + \frac{3}{1}) (1 + \frac{5}{4}) (1 + \frac{7}{9}) \dots (1 + \frac{2 n + 1}{n^{2}}) = (n + 1)^{2}$
Step l :For n=1,i.e., $P (1) = (1 + 1)^{2} = 2^{2} = 4 = (1 + \frac{3}{1})$ which is true.
Step lI: Let it is true for n = k,
$(1 + \frac{3}{1}) (1 + \frac{5}{4}) (1 + \frac{7}{9}) \dots (1 + \frac{2 k + 1}{k^{2}}) = (k + 1)^{2} ...(i)$
Step III For n= k + 1,
$\begin{matrix} \{(1 + \frac{3}{1}) (1 + \frac{5}{4}) (1 + \frac{7}{9}) \dots (1 + \frac{2 k + 1}{k^{2}})\} \{1 + \frac{2 k + 2 + 1}{(k + 1)^{2}}\} \\ = (k + 1)^{2} \{1 + \frac{2 k + 3}{(k + 1)^{2}}\} [using Eq. (i)] \\ = (k + 1)^{2} [\frac{(k + 1)^{2} + 2 k + 3}{(k + 1)^{2}}] = k^{2} + 2 k + 1 + 2 k + 3 \\ = (k + 2)^{2} = [(k + 1) + 1]^{2} \\ [∵ (a + b)^{2} = a^{2} + 2 ab + b^{2}] \end{matrix}$
Therefore, P(k + 1) is true when P(e) is true.
Hence, from the principle of mathematical induction, the statement is true for all natural numbers n.

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