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Before I grind some algebra I was wondering if there was a known equation for a series of the form:

$(x-y)+(x-2y)+(x-3y)+\dots+(x-ny) = T$

Also a variant:

$T-(q+y)-(q+2y)-(q+3y)-\dots-(q+ny) = 0$

The first goes from $0$ to $T$ and the second from $T$ to $0$.

$x$, $y$, and $T$ are known for each instance, but change with each instance.

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    $T$ is not the same in those two expressions, and if they are then $q=x-(n+1)y$2012-05-13

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Answer changed to match corrected problem statement:

It really does just require knowledge of the formula for the triangular numbers:

$\begin{align*} (x-y)+(x-2y)+\ldots+(x-ny)&=nx-y(1+2+\ldots+n)\\ &=nx-y\cdot\frac{n(n+1)}2\\ &=n\left(x-\frac{n+1}2y\right)\;, \end{align*}$

and

$\begin{align*} T-(q+y)-(q+2y)-\ldots-(q+ny)&=T-\Big(nq+y(1+2+\ldots+n)\Big)\\ &=T-\Big(nq+y\cdot\frac{n(n+1)}2\Big)\\ &=T-n\left(q+\frac{n+1}2y\right)\;. \end{align*}$

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    My bad, a typo in the question I failed to notice...2012-05-10