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 $\displaystyle \sum$ symbol, then
$\displaystyle x+2x+3x+4x+5x+6x+7x+8x+9x+10x+11x=\sum_{i=1}^{11}( ix)$
If you want to simplify that expression, then you can use the formula for the sum of an arithmetic progression:
$\displaystyle x+\dots+11x=(1+\dots+11)x=\frac{11\cdot12}{2}x=66x$
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 $\displaystyle \sum_{i=1}^{12} ix$.
Below $\displaystyle \sum$ (capital Greek sigma) we typically write $\displaystyle i = 1$, not just 1, but above $\displaystyle \sum$ we write just 12 (for example). The notation $\displaystyle \sum_{i=1}^{12}$ means the sum where $\displaystyle i$ ranges from 1 to 12.
$\displaystyle \sum_{i=1}^{12}ix$ 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
$\displaystyle 1x+2x+3x+4x+5x+6x+7x=\sum_{i=1}^7ix$. To represent 8x+8x+8x+8x+8x you can write $\displaystyle \sum_{i=8}^{12}8x$, or $\displaystyle \sum_{i=1}^{5}8x$ (note that there is no $\displaystyle i$ after $\displaystyle \sum$, which means that all terms in the sum are the same), or just $\displaystyle 5\cdot 8x$.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 $\displaystyle \sum_{i=7}^5ix$, where the upper limit is less than the lower one.
$\displaystyle \sum_{i=1}^6ix+\sum_{i=1}^58x=\left(\sum_{i=1}^6i+ \sum_{i=1}^58\right)x=(21+40)x=61x$. However, in post #8 the expression labeled 500 has $\displaystyle 1x+2x+3x+4x+5x+6x+7x+8x+8x+8x+8x+8x$ = $\displaystyle \sum_{i=1}^7ix+\sum_{i=1}^58x = 28x+40x = 68x$, which is indeed 680000 for x = 10000.