Probably in the wrong place but Im new so... I need to multiply a number with a constant increase and i dont know how without actually doing the math every time. for instance... 100 x 1000, but adding 10 every time....100, 110, 120, 130, 140, 150, 160 etc. any thoughts? I hate to think i have to add them 1000 times.
multiplication with a constant increase
- Thread starter spooksta
- Start date
yeah spread sheet it boss... you could most likely work out a way to do it but WHY?? Sounds like all you want is the answer so who cares if you didnt have to work to get it..... The computer will do it all for you.
But if you want to know HOW to perfect this process manually, then let me know when you find out how eh.... haha
This sort of thing could work on the spreadsheet - submit formulas to every cell and just move down with
say y = 10
so......
100x1000
(100+Y)x1000
(100+2Y)x1000
Thats prob not the ideal formula basis but its that sort of an arrangement that will give you your weekend back
But if you want to know HOW to perfect this process manually, then let me know when you find out how eh.... haha
This sort of thing could work on the spreadsheet - submit formulas to every cell and just move down with
say y = 10
so......
100x1000
(100+Y)x1000
(100+2Y)x1000
Thats prob not the ideal formula basis but its that sort of an arrangement that will give you your weekend back
better explanation, this again is just an example. I'm more or less just if there is some formula or equation to achieve this. 100x5=500 simple enough but instead of 100+100+100+100+100, what i need to do is 100+110+120+130+140=600. i want to multiply any given number any number of times but with a constant increase at every increment. I just didn't know if there was a way to the math in one equation or if I had to work it out 1 multiple at a time. Follow?
- Joined
- Apr 12, 2005
- Messages
- 11,339
There are a few simple rules for the summation of arithmetic progressions.
Usually:
a = the first in the sequence
d = common difference
Adding up 'n' of them is just this:
a + (a+d) + (a+2d) + ... + (a + (n-1)d) = n*a + d*(1 + 2 + 3 + ... + (n-1)) = n*a + d*[n*(n-1)/2]
That last part may seem a little magic. Feel free to prove it.
n = 3, then 1 + 2 = 3 = (3 * 2 / 2) = 3
n = 7, then 1 + 2 + 3 + ... + 7 = (7 * 6 / 2) = 21
Usually:
a = the first in the sequence
d = common difference
Adding up 'n' of them is just this:
a + (a+d) + (a+2d) + ... + (a + (n-1)d) = n*a + d*(1 + 2 + 3 + ... + (n-1)) = n*a + d*[n*(n-1)/2]
That last part may seem a little magic. Feel free to prove it.
n = 3, then 1 + 2 = 3 = (3 * 2 / 2) = 3
n = 7, then 1 + 2 + 3 + ... + 7 = (7 * 6 / 2) = 21