How does an index greater than 1 affect this arithmatic series?

Question

I was wondering how the starting index affects this arithmatic series and series in general when using the formulas:

$\displaystyle=2\times\left(\frac{41\times40}2\right)-1$

$\displaystyle=1640-1$

$\displaystyle=1639$

if the index starts at 3 instead of say 1, does this affect the answer?

Answer 1

\begin{align} \sum_{i=3}^{40}(2i-1)&=\left(\sum_{i=3}^{40}2i\right)-\left(\sum_{i=3}^{40}1\right)\\ &=2(3+4+\dotsb+40)-(\underbrace{1+1+\dotsb+1}_\textrm{38 of these})\\ &=2\big((1+2+3+4+\dotsb+40)-(1+2)\big)-(\underbrace{1+1+\dotsb+1}_\textrm{38 of these})\\ &=2\Bigg(\left(\frac{40\times41}{2}\right)-(1+2)\Bigg)-(38)\\ &=2\left(\frac{40\times41}{2}-3\right)-(38)\\ &=2(820-3)-38\\ &=1596 \end{align}

Answer 2

I am not sure that I understand exactly what you are asking, but I assume you are looking for something like this. The following holds true: $$ \sum_{k=1}^n k= \frac{n(n+1)}{2}. $$ Now you have $$ \sum_{k=\rho}^n k = \sum_{k=1}^n k- \sum_{k=1}^{\rho-1} k =\frac{n(n+1)}{2} - \frac{\rho(\rho-1)}{2} = \frac{n^2+n-\rho^2+\rho}{2}. $$

Edit: After your clarification, you may use the fact that $$ \sum_{k=1}^n (2k-1)=2\Bigg(\sum_{k=1}^n k\Bigg) - n. $$

Answer 3

Let us calculate the following sum where $2\le m\le 40$ ($m=3$ in your case): $$\begin{align}\sum_{i=\color{red}{m}}^{40}(2i-1)&=\sum_{i=1}^{40}(2i-1)-\sum_{i=1}^{m-1}(2i-1)\\&=2\cdot\frac{40\cdot 41}{2}-40-2\cdot\frac{(m-1)m}{2}+(m-1)\\&=1600-(m-1)^2.\end{align}$$

As you can see, the result contains $m$. This means $m$ does affect the result.

P.S. Your answer is not corrrect. Set $m=3$ in the above result to get $1596$ as the answer.