Video Transcript
We can find a formula for the sum of a geometric series. Consider the series 𝑆 sub 𝑛 equals 𝑎 plus 𝑎𝑟 plus 𝑎𝑟 squared plus 𝑎𝑟 cubed all the way up to 𝑎𝑟 to the power of 𝑛 minus one.
Then there are several parts to this question. We’ll do these bit by bit. The first asks us to multiply the expression for 𝑆 sub 𝑛 by 𝑟, the common ratio.
Our 𝑆 sub 𝑛 is given here. It’s 𝑎 all the way up to 𝑎𝑟 to the power of 𝑛 minus one. We’re going to multiply everything by 𝑟. Now, on the left-hand side, that’s fairly straightforward. We just get 𝑟𝑆 sub 𝑛. Then, on the right-hand side, we get an expression involving a pair of parentheses. It’s 𝑟 times 𝑎 plus 𝑎𝑟 plus 𝑎𝑟 squared and so on. We’re now going to distribute the parentheses on the right-hand side. 𝑟 times 𝑎 we’ll write as 𝑎𝑟. Next, 𝑟 times 𝑎𝑟 is 𝑎𝑟 squared, 𝑟 times 𝑎𝑟 squared is 𝑎𝑟 cubed, and we repeat this process for 𝑎𝑟 cubed.
We also now need to multiply 𝑟 by 𝑎𝑟 to the power of 𝑛 minus one. Well, that’s the same as multiplying 𝑎𝑟 to the power of 𝑛 minus one by 𝑟 to the power of one. But remember, we know that when we’re multiplying two expressions whose bases are equivalent, we can simply add the exponents. So we’re going to add 𝑛 minus one and one, and that leaves us simply with 𝑎𝑟 to the power of 𝑛. So we have multiplied our expression for 𝑆 sub 𝑛 by 𝑟.
Let’s clear some space and look at the second part of this question.
So, we have 𝑆 sub 𝑛 equals 𝑎 plus 𝑎𝑟 all the way up to 𝑎𝑟 to the power of 𝑛 minus one. And 𝑟 times 𝑆 sub 𝑛 equals 𝑎𝑟 plus 𝑎𝑟 squared all the way up to 𝑎𝑟 to the 𝑛th power. The right-hand sides of the equations are very similar. Identify the terms that do not appear on the right-hand side of both equations.
In order to answer this part of the question, let’s begin by identifying the terms that do appear on the right-hand side of both equations. We see there are 𝑎𝑟 in both equations, 𝑎𝑟 squared, and 𝑎𝑟 cubed. In fact, we can infer that 𝑎𝑟 to the fourth power is likely to be in the first equation, and we can also make an assumption that 𝑎𝑟 to the power of 𝑛 minus one must be in our second equation. And that’s because it goes up to the next term, which is 𝑎𝑟 to the power of 𝑛 here. This means we’re left with two terms. In our first equation, we’re left with 𝑎, and in our second, we’re left with 𝑎𝑟 to the 𝑛th power. So, the terms that do not appear on the right-hand side of both equations are 𝑎 and 𝑎𝑟 to the 𝑛th power.
Let’s now clear some space and look at the third part of this question.
Now, consider the subtraction 𝑆 sub 𝑛 minus 𝑟𝑆 sub 𝑛 which is equal to 𝑎 plus 𝑎𝑟 all the way up to 𝑎𝑟 to the power of 𝑛 minus one minus 𝑎𝑟 plus 𝑎𝑟 squared all the way to 𝑎𝑟 to the 𝑛th power. Use the answer to the previous part to simplify the subtraction 𝑆 sub 𝑛 minus 𝑟𝑆 sub 𝑛.
Remember, when we first looked to answer that previous question, we first identified which terms each expression had in common. We now know that when we subtract these two expressions, those terms that appear in both will cancel out. 𝑎𝑟 minus 𝑎𝑟 is zero; 𝑎𝑟 squared minus 𝑎𝑟 squared is zero. And then, of course, we can continue in this manner, and we’ll be left with the terms 𝑎 and 𝑎𝑟 to the 𝑛th power. In fact, because we’re subtracting every term in the second expression, 𝑆 sub 𝑛 minus 𝑟𝑆 sub 𝑛 is going to be 𝑎 minus 𝑎𝑟 to the 𝑛th power.
Then the fourth part of this question asks us to factor both sides of the equation.
So let’s begin by looking at the left-hand side. We notice that each term in this expression has a factor of 𝑆 sub 𝑛. So when we divide 𝑆 sub 𝑛 by 𝑆 sub 𝑛, we get one. And when we divide 𝑟𝑆 sub 𝑛 by 𝑆 sub 𝑛, we simply get 𝑟. Similarly, on the right-hand side, there’s a common factor of 𝑎. When we divide 𝑎 by 𝑎, we get one. And when we divide negative 𝑎𝑟 to the 𝑛th power by 𝑎, we’re left with negative 𝑟 to the 𝑛th power. And so we fully factored both sides of our equation. We get 𝑆 sub 𝑛 times one minus 𝑟 equals 𝑎 times one minus 𝑟 to the 𝑛th power.
Finally, the last part of this question tells us to rearrange the equation to make 𝑆 sub 𝑛 the subject of the formula.
So looking at the previous question, we had 𝑆 sub 𝑛 times one minus 𝑟 equals 𝑎 times one minus 𝑟 to the 𝑛th power. To make 𝑆 sub 𝑛 the subject, it follows then that we’re going to need to divide both sides of this equation by one minus 𝑟. When we do on the left-hand side, we’re simply left with 𝑆 sub 𝑛. Then on the right-hand side, we get 𝑎 times one minus 𝑟 to the 𝑛th power over one minus 𝑟. And so we found the formula for the sum of a geometric series with 𝑛th term 𝑎𝑟 to the power of 𝑛 minus one. It’s 𝑆 sub 𝑛 equals 𝑎 times one minus 𝑟 to the 𝑛th power over one minus 𝑟.
Now at this stage, it’s worth noting that this is a full derivation of the formula for the sum of a geometric series. We don’t need to perform this process every time we’re going to find the sum, but we do need to know where it comes from. And instead, we can just learn the formula 𝑆 sub 𝑛 equals 𝑎 times one minus 𝑟 to the 𝑛th power over one minus 𝑟.
