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.

Guest 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.

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.**

Guest 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).

MaxWong 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!.**

Guest Mar 9, 2019

edited by
Guest
Mar 9, 2019

edited by Guest Mar 9, 2019

edited by Guest Mar 9, 2019

edited by Guest Mar 9, 2019

edited by Guest Mar 9, 2019