Find the formula for $1 + 4 + 7 +\ldots+ (3n - 2)$ and verify the formula by induction

Question

I wrote it as the following sum:

$$1 + \sum_{k=1}^n (3k - 2)$$

Which I solved for and got the following formula:

$$\frac{3n^2 - n + 2}2$$

But this seems wrong to me because the base case seems incorrect to me. Any help with this?

Answer 1

The sum should be $\sum_{k=1}^n (3k-2)$.

Using the arithmetic series formula, we have

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

Try to prove it using induction.

Answer 2

The numbers $1,\,4,\,7,\,\ldots,\,(3n-2)$ form an arithmetic progression. The first term is $a_1=1$, and the $n$-th term is $3n-2$, so their sum, $S$, is given by $$S=\frac{n\left[1+(3n-2)\right]}2=\frac{n(3n-1)}{2}$$