Answer
See explanation
Work Step by Step
For n=1,
$LHS=\Sigma_{i=1}^12i = 2(1) =2$
$RHS = n(n+1) = 1(1+1) =2$
Therefore the formula is true for the base case.
Now let it be true for some n=k.
So, p(k): $\Sigma_{i=1}^k2i= k(k+1)$
Adding 2(k+1) to both sides gives,
$2(k+1)+\Sigma_{i=1}^k2i= k(k+1)+2(k+1)$
$\Sigma_{i=1}^{k+1}2i= (k+1)(k+2)$
$\Sigma_{i=1}^{k+1}2i= (k+1)((k+1)+1)$
So, $p(k)→p(k+1)$ and p(1) is true. Thus, by the principle of mathematical induction, the proposition is true for all natural numbers
