Question

By mathamatical induction $n(n^2-1)$ is divisible by

• A. $19$
• B. $23$
• C. $24$
• D. $29$

Answer 1

The correct option is C $24$

Let $P\left(n\right)=n(n^2-1)$

Step I: For $n=1$,

$P\left(1\right)=1(1^1-1)$

$=$ 0, 0 is divisible by 24.

Therefore, the result is true for $n=1$ .

Step II: Assume that the result is true for $n=k$
i.e., $P\left(k\right)=k(k^2-1)$ is divisible by 24
$\Rightarrow P\left(k\right)=24r$, when r is an integer and k is an odd positive integer.

Step III: For $n=k+2$
$\therefore P(k+2)=(k+2)\left(\right(k+2{)}^2-1)$
$=(k+2)(k^2+4k+3)$
$=k^3+6k^2+11k+6$
$=24r+k+6k^2+11k+6$ (From assumption step)
$=6k^2+12k+6+24r$
$=6(k+1{)}^2+24r$
$=6(2m{)}^2+24r$ ($\because$ k + 1 is even, let k + 1 $=$ 2m)
$=24(m^2+r)$

$\because m^2+r$ is an integer, $\therefore 24(m^2+r)$ is clearly divisible by 24.

Hence $P(k+2)$ is divisible by $24$. This shows that the result is true for $n=k+2$.
Hence by the principle of mathematical induction, the result is true for n odd integer.