How to simplify summation $(1\cdot2) + (2\cdot3) + (3\cdot4) + (4\cdot5) + (5\cdot6) + ... + (N\cdot(N+1))$ in terms of N?

Question

I have function f(n) like:

 N=1   result = 2
 N=2   result = 8
 N=3   result = 20
 N=4   result = 40
 N=5   result = 70
 N=6   result = 112
 N=7   result = 168
 N=8   result = 240
 N=9   result = 330
 N=10  result = 440

I could understand that its something like:

N = 1,   (1 + 1) = 2

N = 2,   (1 + 1) + (3 + 3) = 8

N = 3,   (1 + 1) + (3 + 3) + (4 + 4 + 4)= 20

N = 4,   (1 + 1) + (3 + 3) + (4 + 4 + 4) + (5 + 5 + 5 + 5)  =  40

N = 5,   (1 + 1) + (3 + 3) + (4 + 4 + 4) + (5 + 5 + 5 + 5) + (6 + 6 + 6 + 6 + 6) = 70

N = 6,   (1 + 1) + (3 + 3) + (4 + 4 + 4) + (5 + 5 + 5 + 5) + (6 + 6 + 6 + 6 + 6) + (7 + 7 + 7 + 7 + 7 + 7)= 112

Finally, I could understood that sum of N in three loop is:

[Question]

(1*2) + (3*2) + (4*3) + (5*4) + (6*5) + ... + (N * (N-1))

Now, I wants to simplify this equation in terms of N?

A similar kind of question I have asked previously Here also, If I compare results of both series it looks the result in term of N is = ( ((N) * (N+1) * (N+2)) / 3 ). but I am not sure.

Can someone help me to simplify this question further.

Edit: After Mr.Mark Bennet's comment I recheck my equation it should be actually following (notice the small change):

(1*2) + (2*3) + (3*4) + (4*5) + (5*6) + ... + (N * (N+1))

I am tiring to answer some other problem related to computer science (I am from computer so its hard to solve me I was trying since an hours). Can someone also improve my tags — Grijesh Chauhan, Jun 10 '13 at 11:18
You might do better to write the series as: $(1+1)+(2+2+2)+(3+3+3+3)+(4+4+4+4+4)+\dots$ so that all the terms are in the same form. — Mark Bennet, Jun 10 '13 at 11:24
@Mark Why not make the first term one $2$ instead? That saves a lot of effort. — Lord_Farin, Jun 10 '13 at 11:25
@MarkBennet Got it Thanks!!, But I am sure for my final sequence (1*2) + (3*2) + (4*3) + (5*4) + (6*5) + ... + (N * (N-1)) — Grijesh Chauhan, Jun 10 '13 at 11:30
@Lord_Farin - I did it so the number $1$ was visible in the first term. Of course your suggestion is more efficient. — Mark Bennet, Jun 10 '13 at 11:46
$2\times 1+3 \times 2+4\times 3+ 5\times 4 +\dots$ to put all the terms in the same form ... — Mark Bennet, Jun 10 '13 at 11:48
This post is about very similar sum: http://math.stackexchange.com/questions/707256/how-to-show-sum-i-1n-binomi2-binomn13 — Martin Sleziak, Dec 16 '15 at 14:18

vonbrand · Accepted Answer · 2013-06-10T13:43:20.340

3

By the comment, your sum is: $$ \sum_{1 \le k \le n} k (k + 1) = \sum_{1 \le k \le n} k^2 + \sum_{1 \le k \le n} k = \frac{n (n + 1) (2 n + 1)}{6} + \frac{n (n + 1)}{2} = \frac{n (n + 1) (n + 2)}{3} $$ Or even simpler, with $k^{\overline{m}} = k (k + 1) \ldots (k + m - 1)$, you have $$ \sum_{1 \le k \le n} k^{\overline{m}} = \frac{n^{\overline{m + 1}}}{m + 1} $$ In your case $m = 2$.

edited Jun 10 '13 at 13:43

answered Jun 10 '13 at 11:27

vonbrand

27,812

do you means (1*2) + (3*2) + (4*3) + (5*4) + (6*5) + ... + (N * (N-1)) ? by last description – Grijesh Chauhan Jun 10 '13 at 11:28
Hi please read Edit part its actually: (1*2) + (3*2) + (4*3) + (5*4) + (6*5) + ... + (N * (N+1)) Sorry I mistake I made – Grijesh Chauhan Jun 10 '13 at 11:37
Please check my answer .. – Grijesh Chauhan Jun 10 '13 at 11:52
ok So you updated your answer.. its correct na Thanks a lot vonbrand! – Grijesh Chauhan Jun 10 '13 at 12:11

Grijesh Chauhan · Answer 2 · 2013-06-10T17:12:02.983

1

With the help of Mr.Vonbrand and Mr.Martin Argerami (my previous answer I linked), I could write the simplified question for my edited series:

    (1*2) + (2*3) + (3*4) + (4*5) + (5*6) + ... + (n * (n+1))

Solution:

$$ \sum_{1 \le k \le n} k (k + 1) = \sum_{1 \le k \le n} k^2 + \sum_{1 \le k \le n} k = \frac{n (n + 1) (2 n + 1)}{6} + \frac{n (n + 1)}{2} = \frac{n (n + 1)}{2} (\frac{(2n + 1)}{3} + 1) = \frac{n (n + 1)}{2} (\frac{(2n + 4)}{3} ) = \frac{n(n+1)2(n+2)}6. = \frac{n(n+1)(n+2)}3. $$

Let me know if I am wrong.

edited Jun 10 '13 at 17:12

answered Jun 10 '13 at 11:51

Grijesh Chauhan

325
4
18

1

What I am trying to explain is that some of your terms have the higher number first, and some the lower number (12, 32, 43, N(N+1)). It would help in organising your work, and in recognising patterns, if the terms were all written the same way. – Mark Bennet Jun 10 '13 at 12:00
@MarkBennet Got it!..done thanks :) – Grijesh Chauhan Jun 10 '13 at 12:04
@GrijeshChauhan Not to mention that $n$ suddenly becomes $N$ for no reason... – Jun 10 '13 at 14:41
@ByronSchmuland oh! this because I partially copied from linked question, Now corrected. Thanks :) – Grijesh Chauhan Jun 10 '13 at 17:13

How to simplify summation $(1\cdot2) + (2\cdot3) + (3\cdot4) + (4\cdot5) + (5\cdot6) + ... + (N\cdot(N+1))$ in terms of N?

2 Answers2