Question

Discrete Math help please

Answer 1

Set $n=1$. Then

$\displaystyle\sum_{i=1}^1(3i-1)=3(1)-1=2$
$\dfrac{1(3(1)+1)}2=\dfrac42=2$

Both sides match, so the statement holds for this case.

Assume it holds for $n=k$. Then

$\displaystyle\sum_{i=1}^{k+1}(3i-1)=\sum_{i=1}^k(3i-1)+3(k+1)-1$
$=\dfrac{k(3k+1)}2+3(k+1)-1$
$=\dfrac{k(3k+1)}2+\dfrac{2(3k+2)}2$
$=\dfrac{3k^2+7k+4}2$
$=\dfrac{(k+1)(3k+4)}2$
$=\dfrac{(k+1)(3(k+1)+1)}2$

so that the statement also holds for $n=k+1$, thus proving the statement by induction.