Algebraically solve a component in a partial sum

I would like to solve this equation for y:

$$T = -a + \sum_{1}^{n} \frac{\left(\frac{x}{n} - \frac{y}{n} \right)}{ (1+b)^{n} }$$

The partial sum (Σ) is from 1 to n. I use the ^ symbol for an exponent. For my purposes all the terms are known except y. Therefore I want to solve the equation for y.

Is this possible?

Thanks,

Mike

• Are you sure about the role of $n$ in this equation? It probably should not be both the number of terms ("from $1$ to $n$") and appearing in the sum. Commented Aug 5, 2015 at 2:07
• Eish, well the number of terms is n (this is a number from 1 to 20 generally) and n also appears in the equation. Truth be told this is an algebraic working of an NPV equation where n is the total number of cash flows (years). Therefore I intend the number of terms to be from year 1 to n: NPV = -InitCapex + ( Σ ( (ReserveValue/n) – (Total Operating Cost/n)) / (1+discount)^n))
– Mike
Commented Aug 5, 2015 at 2:24
• if the sum is from $i=1$ to $i=n$, how do your terms depend on $i$? Commented Aug 5, 2015 at 3:03

My interpretation of your formula is $T = -a + \sum_{i=1}^{n}\frac{\frac{x}{i}-\frac{y}{i}}{(1+b)^i}$, with $1+b\neq 0$, and $y$ not depending on $i$. If this is correct, then it should be pretty straightforward to isolate $y$,

$T = -a + \sum_{i=1}^{n}\frac{\frac{x}{i}-\frac{y}{i}}{(1+b)^i} = -a + \sum_{i=1}^{n}\frac{x}{i(1+b)^i} -\sum_{i=1}^{n}\frac{y}{i(1+b)^i} = -a + \sum_{i=1}^{n}\frac{x}{i(1+b)^i} -y\sum_{i=1}^{n}\frac{1}{i(1+b)^i}$.

So rearranging terms,

$-y\sum_{i=1}^{n}\frac{1}{i(1+b)^i} = T + a - \sum_{i=1}^{n}\frac{x}{i(1+b)^i}$

and assuming $\sum_{i=1}^{n}\frac{1}{i(1+b)^i}\neq 0$, we have multiplicative inverses, so we have

$y = \frac{-T - a + \sum_{i=1}^{n}\frac{x}{i(1+b)^i}}{\sum_{i=1}^{n}\frac{1}{i(1+b)^i}}$

• Thank you for the solution! Your interpretation is correct, however you have helped me realise my question! The only difference is that the terms x/i - y/i should be x/n - y/n in your interpretation of my formula. This is because x and y are totals that need to be divided evenly over the life of the sum.
– Mike
Commented Aug 5, 2015 at 11:00