Answer
See proof
Work Step by Step
We have to prove the statement:
$3+6+9+.....+3n=\dfrac{3n}{2}(n+1)$
Step 1: Prove that the statement is true for $n=1$:
The left side is:
$3(1)=3$
The right side is:
$\dfrac{3(1)}{2}(1+1)=\dfrac{3}{2}\cdot 2=3$
As $3=3$, the statement is true for $n=1$.
Step 2: Assume the statement is true for $k$ implies it is true for $k+1$ too.
We have to prove:
$3+6+9+.....+3k+3(k+1)=\dfrac{3(k+1)}{2}((k+1)+1)$
$3+6+9+.....+3k+3(k+1)=\dfrac{3(k+1)}{2}(k+2)$
Because the statement is true for $k$ we have:
$3+6+9+.....+3k=\dfrac{3k}{2}(k+1)$
Add $3(k+1)$ on each side:
$3+6+9+.....+3k+3(k+1)=\dfrac{3k}{2}(k+1)+3(k+1)$
$3+6+9+.....+3k+3(k+1)=\dfrac{3k(k+1)+6(k+1)}{2}$
$3+6+9+.....+3k+3(k+1)=\dfrac{(k+1)(3k+6)}{2}$
$3+6+9+.....+3k+3(k+1)=\dfrac{3(k+1)}{2}(k+2)$
We found that the statement is also true for $k+1$.
So, the given statement is true.