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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.

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  • $\begingroup$ I added the simplification. Thanks. $\endgroup$
    – Infinity
    Apr 11 '14 at 19:14
  • $\begingroup$ 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. $\endgroup$ Apr 11 '14 at 19:20
  • $\begingroup$ 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. $\endgroup$
    – Infinity
    Apr 11 '14 at 19:26
  • $\begingroup$ 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. $\endgroup$ Apr 11 '14 at 19:30
  • $\begingroup$ That's right. Thanks for the hint. $\endgroup$
    – Infinity
    Apr 11 '14 at 19:41
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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.

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

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  • $\begingroup$ I think you mean by the final result I should replace n with k. Am I right? $\endgroup$
    – Infinity
    Apr 11 '14 at 19:17
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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.

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