Hello,
I am trying to create a summation equation from this algebra equation, would it be possible?
x+2x+3x+4x+5x+6x+7x+8x+9x+10x+11x
If so can anyone help me?
Strictly speaking, this is not an equation because it lacks the = sign. It's an expression. If by "summation equation" you mean an expression that uses the symbol, then
If you want to simplify that expression, then you can use the formula for the sum of an arithmetic progression:
Thank you for the response, I understand that in order to create the equation that symbol is needed but my question is would it be possible to convert the equation I posted into that summation? if so how?
basically it works by set periods for example if the period is set to 3 then the equation would be: x+2x+3x and if the period is set to 4 then the equation would be x+2x+3x+4x
Hello,
I wanted to ask if I got these written correctly:
900: 1x+2x+3x+4x+5x+6x+6x+6x+6x+6x+6x+6x
500: 1x+2x+3x+4x+5x+6x+7x+8x+8x+8x+8x+8x
100: 1x+2x+3x+4x+5x+6x+7x+8x+9x+10x+11x+12x
For these would the equation be correct written:
900: S (upper: 5 lower: 1) ix + S (upper:6 lower: 1) 6x where x = 10000
500: S (upper: 5 lower: 7) ix + S (upper:5 lower: 1) 6x where x = 10000
500: S (upper: 12 lower: 1) ix where x = 10000
I don't see the relationship between 900, 500, 100 and the expressions after the colon.
In this forum, you can write [tex]\sum_{i=1}^{12} ix[/tex] to get .
Below (capital Greek sigma) we typically write , not just 1, but above we write just 12 (for example). The notation means the sum where ranges from 1 to 12.
is indeed 1x+2x+3x+4x+5x+6x+7x+8x+9x+10x+11x+12x regardless of the value of x.500: S (upper: 12 lower: 1) ix where x = 10000
. To represent 8x+8x+8x+8x+8x you can write , or (note that there is no after , which means that all terms in the sum are the same), or just .500: S (upper: 5 lower: 7) ix + S (upper:5 lower: 1) 6x where x = 10000
No.
In the explicit sum, 6x occurs 7 times, but in the sigma-expression, it is used only 6 times.
It is not clear what 6x is doing in the sigma-expression and why you have , where the upper limit is less than the lower one.