0

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.

Martin Sleziak
  • 53,687
Infinity
  • 271
  • I added the simplification. Thanks. – Infinity 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. – André Nicolas 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. – Infinity 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. – André Nicolas Apr 11 '14 at 19:30
  • That's right. Thanks for the hint. – Infinity Apr 11 '14 at 19:41
  • See also http://math.stackexchange.com/questions/286312/proof-by-induction-that-sum-i-1n-3i-2-fracn3n-12 and http://math.stackexchange.com/questions/1050814/prove-that-1-4-7-3n-%E2%88%92-2-frac-n3n-%E2%88%92-12 – Martin Sleziak Oct 24 '15 at 05:44
  • It does work! n=0 => 1, n=1 => 5 = 1 + 4, n = 2 => 12 = 1 + 4 + 7, etc. – fleablood Oct 24 '15 at 06:03
  • ...but you didn't do it by induction. (which I hate to criticize you on as direct proofs are more satisfying to me). – fleablood Oct 24 '15 at 06:04

3 Answers3

1

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.

daw
  • 49,113
  • 2
  • 38
  • 76
0

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

user2566092
  • 26,142
0

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.

user137481
  • 2,605