2 + 5 + 8 + . . . + (6n-1) = n(6n+1)
This is what I have so far.
The sum of (3j-1) from j=1 to something I`m not sure of.
Asked
Active
Viewed 360 times
2
-
$3j-1=6n-1$ so $j$ is ? – hhsaffar Dec 10 '13 at 22:40
-
Note that the equation is incorrect: the actual sum of the terms on the lefthand side is $n(6n+1)$. – Brian M. Scott Dec 10 '13 at 22:47
2 Answers
2
Could it just be:
$$\sum_{j=1}^{2n} (3j-1)\;?$$
We test this by letting $n=3$. Then, we see that:
$$\sum_{j=1}^{6} (3j-1) = 2 + 5 +8+11+14+17=57 = (3)[(6)(3)-1]=n(6n+1)$$
Nathan Towle
- 165
-
-
-
Well it can't be just $n$ because then the final term of the sum would be $(3n-1)$, not $(6n-1)$, as the problem indicates. Would it not? – Nathan Towle Dec 10 '13 at 22:43
-
1@bof: That error seems to be present in the original problem statement. – Brian M. Scott Dec 10 '13 at 22:44
-
2
$$\sum\limits_{i=1}^{2n} (3i-1) = n(6n+1)$$
This will give you as terms of the series: \begin{align*}3(1)-1&=2\\3(2)-1&=5\\\ldots\\3(2n)-1&=6n-1\end{align*}
EDIT: the problem statement purported that $\sum_{i=1}^{2n} (3i-1) = n(6n-1)$. I will just prove below that this is wrong, and $\sum_{i=1}^{2n} (3i-1) = n(6n+1)$.
\begin{align*} \sum\limits_{i=1}^{2n} (3i-1) &= \sum\limits_{i=1}^{2n} 3i + \sum\limits_{i=1}^{2n}-1 \\ &= \sum\limits_{i=1}^{2n} 3i + (-2n) \\ &= 3\sum\limits_{i=1}^{2n} i + (-2n) \end{align*} Then we just use the summation formula: $\sum_{i=1}^{N}i = \frac{N(N+1)}{2}$. \begin{align*} \\ 3\sum\limits_{i=1}^{2n} i + (-2n) &= 3\left( \frac{2n(2n+1)}{2}\right) + (-2n) \\ &= 3n(2n+1)-2n \\ &= 6n^2+3n-2n \\ &= 6n^2+n \\ &= n(6n+1) \end{align*}
Newb
- 17,672
- 13
- 67
- 114
-
-
-
-
-
1You might want to add that the equation as given in the problem is incorrect: the righthand side should be $n(6n+1)$. – Brian M. Scott Dec 10 '13 at 22:48
-
@BrianM.Scott Indeed. The sum equates to n(6n+1) and not n(6n-1). Good catch. – user114027 Dec 10 '13 at 22:53
-
-
@user114027 I've also proven the fact that it adds to $n(6n+1)$. You may find this useful. – Newb Dec 10 '13 at 23:03
-
Looks good. (I’d already upvoted.) You could also simply use the formula for the sum of an arithmetic series: there are $2n$ terms, the first and last being $2$ and $6n-1$, so the sum is $$\frac{2n(2+(6n-1))}2=n(6n+1);,$$ – Brian M. Scott Dec 10 '13 at 23:04
-
@BrianM.Scott Oh, that didn't occur to me. I didn't even know that formula. Thanks! – Newb Dec 10 '13 at 23:13