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

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.

• I added the simplification. Thanks. Apr 11 '14 at 19:14
• There is still the problem I referred to earlier, your sum should be $\sum_{k=0}^n(3k+1)$. And if induction is asked for, you need to use another approach. Apr 11 '14 at 19:20
• The problem has two parts. The first part is to find a formula for the sum of the sequence. The second part is to prove it using induction. I think if I replace n with k the problem that you referred to is solved. You are also right about the brackets around the expression. Apr 11 '14 at 19:26
• If we will use induction, then by tradition one goes from the case $n=k$ to the case $n=k+1$. Then one would use something lime $\sum_{i=1}^n$, or $\sum_{i=1}^k$ to avoid confusion. Apr 11 '14 at 19:30
• That's right. Thanks for the hint. Apr 11 '14 at 19:41

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