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.

  • 1
    $\begingroup$ 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

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*}


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.