1. Tardigrade
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  4. For all n ∈ N, (1/1 ⋅ 2 ⋅ 3)+(1/2 ⋅ 3 ⋅ 4)+(1/3 ⋅ 4 ⋅ 5)+⋅s+(1/n(n+1)(n+2)) is equal to

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Q. For all $n \in N$, $\frac{1}{1 \cdot 2 \cdot 3}+\frac{1}{2 \cdot 3 \cdot 4}+\frac{1}{3 \cdot 4 \cdot 5}+\cdots+\frac{1}{n(n+1)(n+2)}$ is equal to

Principle of Mathematical Induction

Solution:

Let the statement $P(n)$ be defined as
$P(n): \frac{1}{1 \cdot 2 \cdot 3}+\frac{1}{2 \cdot 3 \cdot 4} +\frac{1}{3 \cdot 4 \cdot 5}+\cdots $
$ +\frac{1}{n(n+1)(n+2)}=\frac{n(n+3)}{4(n+1)(n+2)}$
Step I For $n=1$,
$P(1): \frac{1}{1 \cdot 2 \cdot 3}=\frac{1(1+3)}{4(1+1)(1+2)}=\frac{4}{4 \times 2 \times 3}=\frac{1}{1 \cdot 2 \cdot 3}$
which is true.
Step II Let it is true for $n=k$.
$\frac{1}{1 \cdot 2 \cdot 3}+\frac{1}{2 \cdot 3 \cdot 4}+\frac{1}{3 \cdot 4 \cdot 5}+\ldots+\frac{1}{k(k+1)(k+2)} $
$=\frac{k(k+3)}{4(k+1)(k+2)} .....$(i)
Step III For $n=k+1$,
$\left(\frac{1}{1 \cdot 2 \cdot 3}+\frac{1}{2 \cdot 3 \cdot 4}+\frac{1}{3 \cdot 4 \cdot 5}+\cdots+\frac{1}{k(k+1)(k+2)}\right)$
$ \left.=\frac{k(k+3)}{4(k+1)(k+2)}+\frac{1}{(k+1)(k+2)(k+3)} \text { [using Eq. (i) }\right] $
$= \frac{k(k+3)^2+4}{4(k+1)(k+2)(k+3)} $
$= \frac{k\left(k^2+9+6 k\right)+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)\left(k^2+5 k+4\right)}{4(k+1)(k+2)(k+3)}=\frac{k^2+5 k+4}{4(k+2)(k+3)} $
$=\frac{k^2+4 k+k+4}{4(k+2)(k+3)}=\frac{k(k+4)+(k+4)}{4(k+2)(k+3)}$
$=\frac{(k+1)(k+4)}{4(k+2)(k+3)}=\frac{(k+1)[(k+1)+3]}{4[(k+1)+1][(k+1)+2]}$
Therefore, $P(k+1)$ is true when $P(k)$ is true. Hence, from the principle of mathematical induction, the statement is true for all natural numbers $n$.