Finding the formula for the sum of a the sequence $1 + 4 + 7 + ... + (3n + 1)$

Question

In the problem below, It is asked to find the formula for the sum of the sequence and then to prove whether it is true or false for all n values using induction.

$$ 1 + 4 + 7 + ... + (3n + 1), \ n\in \Bbb N_0$$

In order to do that I tried to convert it into Sigma notation

$$\sum_{n=0}^k 3n + 1 $$

and then using the rules of sigma notation I came up with

$$\sum_{n=0}^k 3n + 1 = 3\cdot \sum_{n=0}^k n + \sum_{n=0}^k 1$$

and then I replaced it with the following to come to the formula for the sum of the sequence

$$3\cdot\frac{n(n+1)}{2} + (n + 1) = \frac{(n+1)(3n+2)}{2}$$

But it seems to be totally incorrect!

What am I doing wrong. Any help is appreciated.

Answer 1

Your answer looks just fine, although you changed the role that $n$ plays in your formula half-way through the post.

Answer 2

Looks good! You should add brackets in this sum $$ \sum_{n=0}^k (3n+1) $$ to clarify whether the $+1$ summand belongs to the sum.

Answer 3

I think you may be confused with what k should be. If you have m terms, then k = m-1. For example, if you have 3 terms 1, 4 and 7, k should be 2 not 3.