I am trying to help a high school student to solve summation calculation problems. The $i$ value of most answers I found was based on $i=1$ like this:
Now, the $i$ value of our questions is $i=0$ (instead of $i=1$). Is there any ready formula available for us to solve these questions? And what if $i=2$ or other numbers? Thanks for your help!
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$\begingroup$In general, you want to make sure that you start/stop the sum with the same terms. For example, this sum starts from 7 and ends at 20:$$\sum_{i=1}^{14}(i+6).$$If we want $i$ to start from $3$ (say), then we need to adjust the function inside the sum, and the upper limit, so that it still starts from 7 and ends at 20:$$\sum_{i=3}^{16}(i+4).$$In general, for any "shift" $D$, we have$$\sum_{i=a}^b f(i) = \sum_{i=a+D}^{b+D}f(i-D) = \sum_{i=a-D}^{b-D}f(i+D).$$In your specific case however, for 2, 3, and 4 you don't need to change anything, since starting from $i=0$ simply adds $0+$ in front. For number 1, you have$$\sum_{i=1}^nc=\sum_{i=0}^{n-1}c=cn.$$This is true for any $n$, so if you want you can put $n=N+1$ to get$$\sum_{i=0}^N c=c(N+1).$$
$\endgroup$ 1 $\begingroup$Just use $i = 0$ in your image. Then for 2), 3), 4) you just add $0$ to your sum. In 1) you add another c, so it changes the summation but it is easy to think about it. Try it yourself.
For $i \geq 2$ you can write it differently, e.g.$$\sum\limits_{i=2}^n c = \sum\limits_{i=1}^n c - \sum\limits_{i=1}^1 c = cn - c = (n-1)c$$
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