Hello everyone. So in this video, I'd like to talk about summations and products, but first I'd like to review what the notation means.
Let $\{a_n\}$ be a sequence. Recall:
$$\sum_{i=1}^n a_i=a_1+a_2+...+a_n$$and
$$\prod_{i=1}^n a_i=a_1\cdot a_2\cdot...\cdot a_n$$With this in mind, let's answer the following questions.
Question: Write $\displaystyle\sum_{i=3}^{28} i(i+1)^3$ using a summation starting from $1$.
So how do we change the bottom index to start from $0$? Well the idea is you're going to use a new letter. So we're going to define a new letter, let's call it $j$, and we're going to define $j=i-2$.
Now you can rewrite the summation using the same letter but I'm going to use a different letter just for ease of notation.
So now that $j=i-2$, when $i=3$, $j=1$ and when $i=28$, $j=26$. So if I replace $i$ with $j+2$ and write up the summations, let's see what we get.
$$\sum_{i=3}^{28} i(i+1)^3=\sum_{j=1}^{26} (j+2)(j+3)^3$$These two summations are equal. This is what you call a substitution. You could substitute a letter for a different letter, and you can modify your summation accordingly.
Now, for example, you can say that these two things are equal. We can also just expand some of the terms out which might be a good exercise.
Let's call it a "for fun" section. So the sum is going to be $\displaystyle\sum_{i=3}^{28} i(i+1)^3$. So if we start to evaluate this sum, right, the first term is $3(3+1)^3$, I'm just plugging in $3$.
I increment $3$ and I go to $4$, and I get the next term: $4(4+1)^3$. Now you increment $i$ to go to $5$, $5 \leq 28$, and I continue and I continue and I continue.
$$\sum_{i=3}^{28} i(i+1)^3=3(3+1)^3+4(4+1)^3+...+28(28+1)^3$$When's the last term? Well the last term is going to occur when $i=28$ equals 28 so I'm going to get $28(28+1)^3$.
So hopefully we all agree that that's true and then we can do this with the other sum as well. Again, this is just for practice; just to make sure we understand the notation that we can actually do this.
$$\sum_{j=1}^{26} (j+2)(j+3)^3=(1+2)(1+3)^3+(2+2)(2+3)^3+...+(26+2)(26+3)$$Hopefully we agree that those two things are equal. We have $(2+2)$, and $(2+3)$ so $4$ and $5$, those two things are equal, and then we keep going and at every step, even though I didn't show all the steps, $28=26+2$, $(28+1)=(26+3)$. These two sums are actually equal, every term is the same.
Hopefully this gives you a idea of how to change the indices so that you can start it at whatever number you'd like, and this might be important for certain applications.
Question: Write $\displaystyle\sum_{i=k+2}^{3k+2}i$ as a summation from $\displaystyle\sum_{i=k}^{3k}i$ (with other terms).
So let's see that now. So I'm not going to expand this sum out like I did the last one, just for sanity's sake. We should be able to do this just by looking at it, right?
Where do these differ? Now the terms in the second sum, there's a couple of extra terms. We start at $k$, but the first sum started at $k+2$, right? So the second one's going to have a $k$, a $k+1$, $k+2$, $k+3$, all the way up to $3k$, but the $k$ and the $k+1$ terms aren't the same. So if I'm going to include the $k$ and $k+1$ terms in the sum, I'd better remove them.
Right now, if you expand this out, $\displaystyle\sum_{i=k}^{3k}i$, this is a sum from $k$, $k+1$, $k+2$, $k+3$, but now I’m subtracting off $k$ and $k+1$. So I could condense this to a sum starting at $k+2$, so these two things at the bottom end are equal.
Now on the top end though, one has a term $3k+1$ and the other one stops at $3k$. So to account for that, I need to include the $3k+1$ term. If I plug in $3k+1$ in for $i$, I'm just going to get $3k+1$.
$$\sum_{i=k+2}^{3k+2}i=\sum_{i=k}^{3k}i-k-(k+1)+(3k+1)$$You could imagine changing this question slightly. For example, if I made everything squared, instead of $i$ I had $i^2$, then I would instead of just $k$ and $k_1$, I would have squares everywhere.
Modified Question: Write $\displaystyle\sum_{i=k+2}^{3k+2}i^2$ as a summation from $\displaystyle\sum_{i=k}^{3k}i^2$ (with other terms).
Here again, I'm starting from $i=k$ to $3k$ of $i^2$. The $k$th and the $(k+1)$st term, these are extra on this side, right, the first one starts at $k+2$ but the second starts at $k$, so I need to remove those in order to get the equality on the bottom end of these two sums to be equal.
And on the top end, well the first one finishes at $3k+1$ and the second one finishes at $3k$ so I need to add in that extra term that I've lost.
$$\sum_{i=k+2}^{3k+2}i^2=\sum_{i=k}^{3k}i^2-k^2-(k+1)^2+(3k+1)^2$$Now, for example, if I change a $3k+1$ to a $3k+2$, then I would just simply add the extra $3k+2$ term.
And there you have it. These are all ways to manipulate summations, these aren't all of the ways but these are some of the ways to do so. So hopefully this gives you a better understanding of the notation and helps you out for your midterm. Best of luck.