Find Starting Number in Sigma Notation Problem Suppose that $\sum_{k=a}^{106} (2k-3) = 10,001$ . What is the value of a? 
You are not allowed to use trial and error, use algebra instead. 
I know that $a=34$ because of an online calculator and trial and error. How would I be able to find the answer algebraically? Thank you.
$$\sum_{k=a}^{106} (2k-3) = 10,001$$
 A: Setup:
$\sum\limits_{k=a}^{106}(2k-3)=2(\sum\limits_{k=a}^{106}k)+(-3)(\sum\limits_{k=a}^{106}1)$
Now... the following two summation formulae are useful to know:
$$\sum\limits_{k=1}^n 1 = n$$
$$\sum\limits_{k=1}^n k = \frac{n(n+1)}{2}$$
With these in mind, we can come up with more general formulae for if the index begins at a different number instead of at $1$ by remembering $\sum\limits_{k=1}^n f(k) = \sum\limits_{k=1}^{a-1}f(k) + \sum\limits_{k=a}^nf(k)$, that is to say we can split up the summation into multiple parts.
Now, we should have a way to express the original expression without the summation symbols as a polynomial in terms of $a$.  We are told that the original expression should be equal to $10001$, so all that remains is to solve for the value of $a$.

Further details:

 For $a\leq n$ we have $\sum\limits_{k=a}^n 1 = \sum\limits_{k=1}^n 1 - \sum\limits_{k=1}^{a-1}1 = n - (a-1) = n-a+1$.  Similarly we have $\sum\limits_{k=a}^n k = \sum\limits_{k=1}^n k - \sum\limits_{k=1}^{a-1} k = \frac{n(n+1)}{2}-\frac{(a-1)a}{2}$

Now... for your specific problem this means:

 $\sum\limits_{k=a}^{106} (2k-3)=2(\frac{106(107)}{2}-\frac{(a-1)a}{2}) - 3(106-a+1)=-a^2+4a+11021$

Which, remember we wanted to be equal to $10001$, giving us the equation to solve:

 $-a^2+4a+11021=10001$.  Now, we subtract $10001$ from each side...

