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# Difficult sequence

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Sum up the following sequence:
1.2 + 1.3 + 1.4 + 1.5+......+99.100 + 2.3 + 2.4 + 2.5+......+99.100 + 3.4 + 3.5 + 3.6+......+99.100.....and so on till...98.99 + 98.100 + 99.100. Please show steps.
Thank you for any help.

Mar 6, 2019

#1
-2

This sequence is too long. Almost impossible to do. Also, all of the information isn't there so it's unsolvable.

Mar 7, 2019
edited by EmeraldWonder  Mar 7, 2019
#2
+1

Well, it is a difficult sequence, or rather sequences! There are actually 99 sequences that you have to sum up individually, and then take the sum of ALL the sums!.

They go like this:
2 + 3 + 4 + 5........100 - with 99 terms and comm. diff=1
6 + 8 + 10 + 12....100 - with 98 terms and comm. diff=2
12 + 15 + 18 +......100 - with 97 terms and comm. diff=3
And so on till: 99 x 100 =9900 - being the sum of the last sequence.
Only a spreadsheet or a computer code could sum them up for you. I wrote a short computer code that sums them up as follows:
5049 +  10094 +  15132 +  20160 +  25175 +  30174 +  35154 +  40112 +  45045 +  49950 +  54824 +  59664 +  64467 +  69230 +  73950 +  78624 +  83249 +  87822 +  92340 +  96800 +  101199 +  105534 +  109802 +  114000 +  118125 +  122174 +  126144 +  130032 +  133835 +  137550 +  141174 +  144704 +  148137 +  151470 +  154700 +  157824 +  160839 +  163742 +  166530 +  169200 +  171749 +  174174 +  176472 +  178640 +  180675 +  182574 +  184334 +  185952 +  187425 +  188750 +  189924 +  190944 +  191807 +  192510 +  193050 +  193424 +  193629 +  193662 +  193520 +  193200 +  192699 +  192014 +  191142 +  190080 +  188825 +  187374 +  185724 +  183872 +  181815 +  179550 +  177074 +  174384 +  171477 +  168350 +  165000 +  161424 +  157619 +  153582 +  149310 +  144800 +  140049 +  135054 +  129812 +  124320 +  118575 +  112574 +  106314 +  99792 +  93005 +  85950 +  78624 +  71024 +  63147 +  54990 +  46550 +  37824 +  28809 +  19502 +  9900 = 12,582,075.
Alternatively, there is a pattern called "closed form" that applies to every term above and it looks like this:
5050*n - 1/2 * [n^3 + n^2]. Example: First term:
5050*1 - 1/2 * [1^3 + 1^2] =5050 - 1 =5049. And with this "closed form" you can use any calculator that has sigma(∑) built into it and sum them all up. You should get the same total as above:12,582,075.

Mar 7, 2019
#3
0

There are 99 sequences to sum up.

The n-th sequence looks like this:

n(n+1) + (n+1)(n+2) + ... + 99(100).

We can solve this by summation:

$$\quad n(n+1 ) + (n+1)(n+2) + \cdots + 99(100)\\ =(1\cdot2 + 2\cdot 3 + ... + 99(100)) - (1\cdot 2 + 2\cdot 3 + ... + n(n-1))\\ =\displaystyle \sum^{99}_{k=1} (k^2 + k) - \sum^{n-1}_{k=1}(k^2 + k)\\ = \dfrac{99(100)(199)}{6} + \dfrac{99(100)}{2} - \dfrac{n(n-1)(2n-1)}{6} - \dfrac{n(n-1)}{2}\\ = -\dfrac{n^3}{3} + \dfrac{n}{3} + 333300$$

Now we need to find the sum of this expression from n = 1 to p, where p is some positive integer.

$$\displaystyle\sum^{p}_{n=1} \left(\dfrac{-n^3}{3} + \dfrac{n}{3} + 333300\right)\\ = \dfrac{p(-p^3-2p^2+p+3999602)}{12}$$

We sum this from p = 1 to p = 100 and we get 1,508,082,510

:)

To Guest: The first sequence is (2 + 3 + 4 + ... + 100) + (6 + 8 + 10 + ... + 200) + ... + (98 * 99 + 98*100 + 99*100).

Mar 9, 2019
#4
0

NO Max! Even though I wrote the last terms as 100s, but they are added up as follows:

(1*2) + (1*3) + (1*4) +..................+(1*100) - Notice that the first term is 2 NOT 1. For 99 terms. Then it continues...

(2*3) + (2*4) + (2*5) +..................+(2*100) - where the first term is 6 for 98 terms. Then it continues.......

(3*4) + (3*5) + (3*6) +..................+(3*100) - where the first term is 12 for 97 terms. And so on for a total of 99 sequences. Notice the sum of the sequences listed above:

5,049, 10,094, 15132.......etc. They are all correct with  last terms as 100, 200, 300.......9900.

The total I got for 12,582,075 is correct.

NOTE: If you divide your total of 1,508, 082,510 / 99 =15,233,157 per sequence!, which is IMPOSSIBLE!. The biggest total is of the 58th sequence with a total of: 193,662!.

Mar 9, 2019
edited by Guest  Mar 9, 2019
edited by Guest  Mar 9, 2019
edited by Guest  Mar 9, 2019