Question

Prove by the Principle of Mathematical Induction: $2+4+6+...+2n=n^2+n$ for all natural numbers n.

Answer 1

Assume given statement

Let given statement be

$P\left(n\right),i.e.,P\left(n\right):2+4+6+...+2n=n^2+n\forall nϵN$

Check that statement is true for $n=1$

We observe that P(n) is true for $n=1,$

Since, $2=1^2+1=2$

Assume P(k) to be true and then prove $P(k+1)$ is true.

Assume that the statement is true for some $n=k$

$\therefore 2+4+6+...+2k=k^2+k\dots \left(1\right)$

To prove:$2+4+6+...+2k+2(k+1)=(k+1{)}^2+(k+1)$

LHS$=2+4+6+...+2k+2(k+1)$

LHS $=k^2+k+2(k+1)$

$=k^2+k+2k+2$

$=k^2+2k+1+k+1$

$=(k+1{)}^2+(k+1)=RHS$

Thus, $P(k+1)$ is true whenever P(k) is true.

Hence, By Principle of mathematical Induction P(n) is true for all natural numbers n.