First slide

Introduction to P.M.I

Question

By mathematical induction $\frac{1}{1 \cdot 2 \cdot 3} + \frac{1}{2 \cdot 3 \cdot 4} + \dots + \frac{1}{n (n + 1) (n + 2)}$

Easy

A

$\frac{n (n + 1)}{4 (n + 2) (n + 3)}$
B

$\frac{n (n + 3)}{4 (n + 1) (n + 2)}$
C

$\frac{n (n + 2)}{4 (n + 1) (n + 3)}$
D

None of these

Solution

Let
$\begin{aligned} P (n) : \frac{1}{1 \cdot 2 \cdot 3} + \frac{1}{2 \cdot 3 \cdot 4} + \dots + \frac{1}{n (n + 1) (n + 2)} \\ = \frac{n (n + 3)}{4 (n + 1) (n + 2)} \end{aligned}$
Step l: For n=1
$LHS = \frac{1}{1.2.3} = \frac{1}{6} and RHS = \frac{1 (1 + 3)}{4 (1 + 1) (1 + 2)} = \frac{1}{6}$ ,P(1 ) is true.
Step II: Let P(k) is true, then
$\begin{aligned} P (k) : \frac{1}{1 \cdot 2 \cdot 3} + \frac{1}{2 \cdot 3 \cdot 4} + \dots + \frac{1}{k (k + 1) (k + 2)} \\ = \frac{k (k + 3)}{4 (k + 1) (k + 2)} \end{aligned}$
Step lll : For n=n+1,then
$\begin{matrix} P (k + 1) : \frac{1}{1 \cdot 2 \cdot 3} + \frac{1}{2 \cdot 3 \cdot 4} + \dots + \frac{1}{k (k + 1) (k + 2)} + \frac{1}{(k + 1) (k + 2) (k + 3)} \\ = \frac{(k + 1) (k + 4)}{4 (k + 2) (k + 3)} \end{matrix}$
$\begin{array}{l} LHS = \frac{1}{1 \cdot 2 \cdot 3} + \frac{1}{2 \cdot 3 \cdot 4} + \dots + \frac{1}{k (k + 1) (k + 2)} \\ = \frac{k (k + 3)}{4 (k + 1) (k + 2)} + \frac{1}{(k + 1) (k + 2) (k + 3)} \\ = \frac{k (k + 3)^{2} + 4}{4 (k + 1) (k + 2) (k + 3)} \\ = \frac{k^{3} + 6 k^{2} + 9 k + 4}{4 (k + 1) (k + 2) (k + 3)} \\ = \frac{(k + 1)^{2} (k + 4)}{4 (k + 1) (k + 2) (k + 3)} = \frac{(k + 1) (k + 4)}{4 (k + 2) (k + 3)} = RHS \end{array}$
Hence, P(k + 1) is true. Hence, by principle of mathematical induction for all n $\in$ N, P(n) is true.

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