Question

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

• A. $\frac{(n+3)}{4(n+1)(n+2)}$
• B. $\frac{n(n+3)}{4(n+1)(n+2)}$
• C. $\frac{4(n+1)(n+2)}{n(n+3)}$
• D. None of these

Answer 1

Answer: (B)

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 $n=k$.
$\frac{1}{1\cdot 2\cdot 3}+\frac{1}{2\cdot 3\cdot 4}+\frac{1}{3\cdot 4\cdot 5}+\dots +\frac{1}{k(k+1)(k+2)}$
$=\frac{k(k+3)}{4(k+1)(k+2)}.....$(i)
Step III For $n=k+1$,
$\left(\frac{1}{1\cdot 2\cdot 3}+\frac{1}{2\cdot 3\cdot 4}+\frac{1}{3\cdot 4\cdot 5}+\cdots +\frac{1}{k(k+1)(k+2)}\right)$
$=\frac{k(k+3)}{4(k+1)(k+2)}+\frac{1}{(k+1)(k+2)(k+3)}\text{ [using Eq. (i) }]$
$=\frac{k(k+3{)}^2+4}{4(k+1)(k+2)(k+3)}$
$=\frac{k\left(k^2+9+6k\right)+4}{4(k+1)(k+2)(k+3)}$
$=\frac{k^3+6k^2+9k+4}{4(k+1)(k+2)(k+3)}$
$=\frac{(k+1)\left(k^2+5k+4\right)}{4(k+1)(k+2)(k+3)}=\frac{k^2+5k+4}{4(k+2)(k+3)}$
$=\frac{k^2+4k+k+4}{4(k+2)(k+3)}=\frac{k(k+4)+(k+4)}{4(k+2)(k+3)}$
$=\frac{(k+1)(k+4)}{4(k+2)(k+3)}=\frac{(k+1)[(k+1)+3]}{4[(k+1)+1][(k+1)+2]}$
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 $n$.