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So, as the title states, I'm trying to make a single summation or integral that will solve for the sequential sums of an already defined algebraic equation. The equation in question is this: $$y=2x+101$$ It's been a few years since I've had to deal with questions like this, so I'm very rusty. In essence, the summation should do something along the lines of this: $$y=101+103+105+107+...+(n-4)+(n-2)+n$$ However, I do want to set up the formula in such a way that I can start anywhere in the sequence, for instance like: $$y=151+153+...+175+177$$ For this reason, I've asked for summation or integral.
At one point, I thought I had it using $y=\int_a^n(2x+101)dx$, but that didn't work the way I wanted. It was 5 off, either way I went.

So, as it stands, I'm a bit stuck. Any help would be greatly appreciated.

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    $\begingroup$ So, I did some more research and found out the problem I was having was an overly basic one. Apparently, I wasn't using the same variable under sigma as I was in the formula, which is what was throwing off the calculations. Sorry to have wasted anyone's time, although I don't know how to close this question. $\endgroup$ – Obsidian Otter May 30 '18 at 12:03
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We recall the summation formula $\sum_{k=1}^n k=\frac{1}{2}n(n+1)$.

We obtain for non-negative integers $0\leq m \leq n$ \begin{align*} \color{blue}{\sum_{k=m}^n(2k+1)}&=2\sum_{k=m}^nk+(n-m+1)\\ &=2\left(\sum_{k=1}^nk-\sum_{k=1}^{m-1}k\right)+(n-m+1)\\ &=2\left(\frac{1}{2}n(n+1)-\frac{1}{2}(m-1)m\right)+n-m+1\\ &\color{blue}{=n^2+2n-m^2+1} \end{align*}

Example: \begin{align*} \color{blue}{151+153+\cdots+177}=\sum_{k=75}^{88}(2k+1)=88^2+2\cdot 8-75^2+1\color{blue}{=1\,296} \end{align*}

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