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