Question

Prove that $n^2-n$ in divisible by $2$ for any positive length $n$.

Answer 1

Let, $f\left(n\right)=n^2-n$; $n\in N$

Now, $f\left(1\right)=0$ is divisible by ${}^{\prime }{2}^{\prime }$

Again $f\left(2\right)=2^2-2=2$; divisible by ${}^{\prime }{2}^{\prime }$

So, $f\left(1\right),f\left(2\right)$ are true.

Let us assume that $f\left(k\right)$ is true i.e.,

$f\left(k\right)=k^2-k=2k_1$; $k_1\in z$

Now,

$f(k+1)=(k+1{)}^2-(k+1)$

$=k(k+1)$

$=k^2+k$

$=k^2-k+2k$

$=2k_1+2k$ [using $\left(1\right)$]

$=2(k_1+k)$;

So, $f(x+1)$ is divisible by $2$.

Now, by principle of mathematical induction we have,

$f\left(x\right)$ is true $\forall n\in N$.