Question

Discrete Math help please

Answer 1

Set
`n=1`. Then


`\displaystyle\sum_(i=1)^1(3i-1)=3(1)-1=2`

`\frac{1(3(1)+1)}2=\frac42=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`

`=\frac{k(3k+1)}2+3(k+1)-1`

`=\frac{k(3k+1)}2+\frac{2(3k+2)}2`

`=\frac{3k^2+7k+4}2`

`=\frac{(k+1)(3k+4)}2`

`=\frac{(k+1)(3(k+1)+1)}2`

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