Question

a(a^2+2)/3 is an integer prove it by induction​

Answer 1

$\dfrac{a(a^2+2)}{3}$

It is a fraction. To get it as an integer, the numerator of the fraction should be exactly divisible by the denominator, i.e., $a(a^2+2)$ should be exactly divisible by 3.

Hence we have only to prove that,

$\large \textsf{ a(a^2+2) is exactly divisible by 3 for any a\geq 1 }$

Let $a=1$

$a(a^2+2)=1(1^2+2)=1(1+2)=1\times 3=3\\ \\ \textsf{Here,}\ \ 3 \mid\ a(a^2+2)\\ \\ \therefore\ \dfrac{a(a^2+2)}{3}\ \ \texttt{is an integer.}$

Let $a=k$

Assume that $3 \mid\ k(k^2+2)$

Let $a=k+1$

$\begin{aligned}&a(a^2+2)\\ \\ \Longrightarrow\ \ &(k+1)((k+1)^2+2)\\ \\ \Longrightarrow\ \ &(k+1)(k^2+2k+1+2)\\ \\ \Longrightarrow\ \ &(k+1)(k^2+2+2k+1)\\ \\ \Longrightarrow\ \ &(k+1)(k^2+2)+(k+1)(2k+1)\\ \\ \Longrightarrow\ \ &k(k^2+2)+k^2+2+(k+1)(2k+1)\\ \\ \Longrightarrow\ \ &k(k^2+2)+k^2+2+2k^2+3k+1\end{aligned}$

$\begin{aligned}\Longrightarrow\ \ &k(k^2+2)+3k^2+3k+3\\ \\ \Longrightarrow\ \ &k(k^2+2)+3(k^2+k+1)\end{aligned}$

Here, we have assumed earlier that $3\mid\ k(k^2+2).$ To this, $3(k^2+k+1),$ which is also a multiple of 3, is added.

$\therefore\ 3\mid\ a(a^2+2)\\ \\ \Longrightarrow\ \dfrac{a(a^2+2)}{3}\ \ \textsf{is an integer.}$

Hence Proved!!!