Suppose that ∑k=1^n ak=0 and ∑k=1^n bk=1 . Find the values of A. ∑i=1^n 8 ak B. ∑i=1^n 250 bi C.

step 1 Good day, ladies and gentlemen. This is section 5 .2, problem number 18. And again, we're given

Suppose that ∑k=1^n ak=0 and ∑k=1^n bk=1 . Find the values...

Key Concepts

Linearity of Summation

The linearity property of summation states that the sum of a sum is the sum of the individual sums, and that constants can be factored out. In other words, for any sequences and constants, the sum of a constant times each term is equal to that constant times the sum of the terms, and the sum of the sum (term1 + term2) is equal to the sum of term1 plus the sum of term2. This principle is especially useful in breaking down complex summations into simpler, more manageable parts.

Summation of Constants

When summing a constant over a specified number of terms in a sequence, the result is simply that constant multiplied by the number of terms. This concept is fundamental in evaluating sums where each term is a constant value, and it simplifies calculations by reducing the summation process to a multiplication operation.

Scalar Multiplication in Summations

Scalar multiplication in summations refers to the property that multiplying each term in a sequence by a constant before summing is equivalent to first summing the sequence and then multiplying by that constant. This property, which is a direct consequence of the linearity of summation, streamlines calculations by allowing us to deal with the scalar separately from the sequence being summed.

Transcript

00:01 Good day, ladies and gentlemen.

00:04 This is section 5 .2, problem number 18.

00:09 And again, we're given two different sums, one over ak and one over bk, and then we're asked to evaluate a collection of sum.

00:23 So here we go.

00:26 Now, in part a, we're asked to evaluate this guy, the sum from 1 to n of 8a k.

00:38 And the first thing i want to point out is that using the constant multiple rule, we can rewrite this guy to equal to 8 times the sum from k equals, whoops, the nice little rule here, of course, from 1 to n of a .k.

01:10 And again, now this is just the constant multiple rule.

01:16 And we know that from our assumption that the inner sum is, or this sum over a, over, over, over, is z.

01:30 So we get that this is equal to 8 times 0, which of course is just equal to 0.

01:45 Okay, so that's our first one.

01:49 And now let's move on to our second one here.

01:56 And our second one is over bk this time, and it is equal to the second one is 250 bk.

02:19 And again, we're going to really, we're just going to do the same type of trick here.

02:25 I mean, it's the same, it's the exact same rule.

02:29 Sorry, the exact same rule, we can just factor out the, or sorry, not factor out, but use the constant multiple rule.

02:43 So this is equal to 250 times, again, the, just as we did before, this is just 250.

02:59 And this time it's of course bk and this sum from my here we get this is one so consequently we get 250 here so this is just equals to 250.

03:18 Well 250 times one but of course that's just two hundred and fifty.

03:23 Okay continuing onward we're going to move to c and this time we're looking at a very similar one but now we're looking at ak this is the sum ak plus plus one okay so now we're looking at this sum here i'll just that out and move the bks, of course, over here.

04:07 Okay, and now the first step, again, we want to apply one of the basic rules of sums, which is the sum, which is what they call the sum rule, meaning that we can split this into the sum over ak plus plus the sum over 1...

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