Question

Statement I For all $n\in N$, $1\cdot 2+2\cdot 3+3\cdot 4+\cdots +n\cdot (n+1)=\frac{n(n+1)(n+2)}{3}$
Statement II For all $n\in N$,
$1\cdot 3+3\cdot 5+5\cdot 7+\cdots +(2n-1)(2n+1)$
$=\frac{n\left(4n^2+6n-1\right)}{3}$

• A. Only Statement I is true
• B. Only Statement II is true
• C. Both statements are true
• D. Both statements are false

Answer 1

Answer: (C)

I. Let the statement $P(n)$ be defined as
$P(n):1\cdot 2+2\cdot 3+\dots +n(n+1)=\frac{n(n+1)(n+2)}{3}$
Step I For $n=1$,
$P(1):1\cdot 2=\frac{1(1+1)(1+2)}{3}=\frac{1\times 2\times 3}{3}=1\cdot 2$
which is true.
Step II Let it is true for $n=k$.
i.e., $1\cdot 2+2\cdot 3+3\cdot 4+\dots +k(k+1)$
$=\frac{k(k+1)(k+2)}{3}....$(i)
Step III For $n=k+1$,
$[1\cdot 2+2\cdot 3+\dots +k(k+1)]+(k+1)(k+2)$
$=\left[\frac{k(k+1)(k+2)}{3}\right]+(k+1)(k+2)\text{ [using Eq. (i) }]$
$=\frac{k(k+1)(k+2)+3(k+1)(k+2)}{3}$
$=\frac{(k+1)(k+2)(k+3)}{3}\text{ [taking }(k+1)(k+2)\text{ [common in numerator part }]$
$=\frac{(k+1)[(k+1)+1][(k+1)+2]}{3}$
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 $n$.
II. Let the statement $P(n)$ be defined as
$P(n):1\cdot 3+3\cdot 5+5\cdot 7+\dots +(2n-1)(2n+1)$
$=\frac{n\left(4n^2+6n-1\right)}{3}$
Step I For $n=1$, we have
$P(1):1\cdot 3=\frac{1\left(4\times 1^2+6\times 1-1\right)}{3}$
$=\frac{4+6-1}{3}=\frac{9}{3}=3=1\cdot 3$
which is true.
Step II Let it is true for $n=k$,
i.e., $1\cdot 3+3\cdot 5+5\cdot 7+\dots +(2k-1)(2k+1)$
$=\frac{k\left(4k^2+6k-1\right)}{3}$
Step III For $n=k+1$,
$[1\cdot 3+3\cdot 5+5\cdot 7+\dots +(2k-1)(2k+1)]$
$+(2k-1+2)(2k+1+2)$
$=\frac{k\left(4k^2+6k-1\right)}{3}+(2k+1)(2k+3)[\text{ using Eq. (i) }]$
$=\frac{k\left(4k^2+6k-1\right)+3\left(4k^2+6k+2k+3\right)}{3}$
$=\frac{4k^3+6k^2-k+12k^2+24k+9}{3}$
$=\frac{4k^3+18k^2+23k+9}{3}$
$=\frac{(k+1)\left(4k^2+14k+9\right)}{3}($factorising it by using factor theorem $)$
$=\frac{(k+1)\left(4k^2+8k+4+6k+5\right)}{3}$
$=\frac{(k+1)\left[4\left(k^2+2k+1\right)+6k+6-1\right]}{3}$
$=\frac{(k+1)\left[4(k+1{)}^2+6(k+1)-1\right]}{3}$
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 $n$.