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.
+219 This sequence is too long. Almost impossible to do. Also, all of the information isn't there so it's unsolvable.
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.
+9519 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).
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!.
