Proof by induction that $\sum_{i=1}^{n}(3i-2)=\frac{n}{2}(3n-1)$

Question

I am trying to perform this proof but I find myself stuck

Prove for all natural number n.

$\sum_{i=1}^{n}(3i-2)=\frac{n}{2}(3n-1)$

The first step ofcourse is P(1) because 1 is the first natural number.

$1=3-2$

Second step let $n=k$ the ass umption step. $\sum_{i=1}^{k}(3i-2)=\frac{k}{2}(3k-1)$

Third step What you try to show.

$P(k+1)$

$\sum_{i=1}^{k+1}(3i-2)=\frac{k+1}{2}(3(k+1)-1)$ The proof

$\sum_{i=1}^{k}(3i-2)= \frac{3(k+1)-2}{2}+**\sum_{i=1}^k(3i-2)$**

That is the sum of the final term plus the sum of all previus terms.

Using the induction assumption

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

$=\frac{3k^2+2k+1}{2}$

But I am not sure how to make it seem like the right hand side in my proof.

Answer 1

The structure of the proof is fine, but there is a small slip in the calculation. We have $$\sum_1^{k+1} (3i-2)=\left(\sum_1^k (3i-2)\right) +3(k+1)-2.$$ By the induction assumption, the right-hand side above is equal to $$\frac{k}{2}(3k-1)+3k+1.$$ Bring to a common denominator and simplify a bit. We get $$\frac{3k^2+5k+2}{2}.$$ The numerator is $(k+1)(3k+2)$, which is $(k+1)(3(k+1)-1)$.