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+(2 n-1)(2 n+1)$
$=\frac{n\left(4 n^2+6 n-1\right)}{3}$

• 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
$P(n): 1 \cdot 2+2 \cdot 3+\ldots+n(n+1)=\frac{n(n+1)(n+2)}{3}$
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+\ldots+k(k+1) $
$=\frac{k(k+1)(k+2)}{3} ....$(i)
Step III For $n=k+1$,
$ {[1 \cdot 2+2 \cdot 3+\ldots+k(k+1)]+(k+1)(k+2) }$
$= {\left.\left[\frac{k(k+1)(k+2)}{3}\right]+(k+1)(k+2) \text { [using Eq. (i) }\right] } $
$= \frac{k(k+1)(k+2)+3(k+1)(k+2)}{3}$
$= \left.\frac{(k+1)(k+2)(k+3)}{3} \text { [taking }(k+1)(k+2) \text { [common in numerator part }\right] $
$= \frac{(k+1)[(k+1)+1][(k+1)+2]}{3}$
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+\ldots+(2 n-1)(2 n+1) $
$ =\frac{n\left(4 n^2+6 n-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., $1 \cdot 3+3 \cdot 5+5 \cdot 7+\ldots+(2 k-1)(2 k+1)$
$ =\frac{k\left(4 k^2+6 k-1\right)}{3} $
Step III For $n=k+1$,
${[1 \cdot 3+3 \cdot 5+5 \cdot 7+\ldots+(2 k-1)(2 k+1)]}$
$+(2 k-1+2)(2 k+1+2)$
$= \frac{k\left(4 k^2+6 k-1\right)}{3}+(2 k+1)(2 k+3) [\text { using Eq. (i) }] $
$= \frac{k\left(4 k^2+6 k-1\right)+3\left(4 k^2+6 k+2 k+3\right)}{3} $
$= \frac{4 k^3+6 k^2-k+12 k^2+24 k+9}{3} $
$= \frac{4 k^3+18 k^2+23 k+9}{3}$
$=\frac{(k+1)\left(4 k^2+14 k+9\right)}{3} ( $factorising it by using factor theorem $)$
$=\frac{(k+1)\left(4 k^2+8 k+4+6 k+5\right)}{3}$
$ =\frac{(k+1)\left[4\left(k^2+2 k+1\right)+6 k+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 $n$.