Video: Finding the General Term in the Binomial Expression

Consider the binomial expansion for (1 + (1/𝑛))^(𝑛). Which of the following expressions is its fourth term? [A] ((1 βˆ’ (1/𝑛))(1 + (2/𝑛)))/3! [B] ((1 βˆ’ (1/𝑛))(1 βˆ’ (2/𝑛)))/3!(𝑛 βˆ’ 3)! [C] (1 βˆ’ (1/𝑛))(1 βˆ’ (2/𝑛)) [D] ((1 βˆ’ (1/𝑛))(1 βˆ’ (2/𝑛)))/3! [E] 𝑛!/3! What is the limit of the (π‘˜ + 1)α΅—Κ° term as 𝑛 tends to infinity? Hence, write in summation (or sigma) notation a series which is equal to the limit of (1 + (1/𝑛))^(𝑛) as 𝑛 tends to infinity. What is the value of this series?

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Video Transcript

Consider the binomial expansion for one plus one over 𝑛 all raised to the 𝑛th power. Which of the following expressions is its fourth term? Option A) One minus one over 𝑛 multiplied by one plus two over 𝑛 all divided by three factorial? Option B) One minus one over 𝑛 multiplied by one minus two over 𝑛 all divided by three factorial multiplied by 𝑛 minus three factorial? Option C) One minus one over 𝑛 multiplied by one minus two over 𝑛? Option D) One minus one over 𝑛 multiplied by one minus two over 𝑛 all divided by three factorial? Or option E) 𝑛 factorial divided by three factorial? What is the limit of the π‘˜ plus oneth term as 𝑛 tends to ∞? Hence, write in summation, or sigma, notation a series which is equal to the limit of one plus one over 𝑛 all raised to the 𝑛th power as 𝑛 tends to ∞. What is the value of this series?

The question is split into parts. The first part of this question is asking us to find the fourth term in the binomial expansion of the expression one plus one over 𝑛 all raised to the 𝑛th power. We recall that the binomial expansions of π‘₯ plus 𝑦 all raised to the 𝑛th power is equal to the sum from π‘Ÿ equals zero to 𝑛 of 𝑛 choose π‘Ÿ multiplied by π‘₯ to the power of 𝑛 minus π‘Ÿ multiplied by 𝑦 to the π‘Ÿth power.

It’s worth noting that we call the summand 𝑛 choose π‘Ÿ multiplied by π‘₯ to the power of 𝑛 minus π‘Ÿ multiplied by 𝑦 to the π‘Ÿth power the π‘Ÿ plus oneth term. This is because π‘Ÿ starts at zero. So, to find the fourth term of this binomial expansion, we’ll want to set π‘Ÿ equal to three. And from the question, we’ll set π‘₯ equal to one and 𝑦 equal to one divided by 𝑛. Substituting these values into our summand gives us 𝑛 choose three multiplied by one to the power of 𝑛 minus three multiplied by one over 𝑛 all cubed.

We now recall that 𝑛 choose π‘Ÿ is shorthand notation for 𝑛 factorial divided by π‘Ÿ factorial multiplied by 𝑛 minus π‘Ÿ factorial. We can use this to rewrite 𝑛 choose three as 𝑛 factorial divided by three factorial multiplied by 𝑛 minus three factorial. We can simplify this further by noticing that one to the power of 𝑛 minus three is just equal to one. And one over 𝑛 all cubed is the same as one over 𝑛 cubed.

We now want to start cancelling shared factors in the numerator and the denominator. We start by noticing that 𝑛 factorial is equal to 𝑛 multiplied by 𝑛 minus one multiplied by 𝑛 minus two multiplied by 𝑛 minus three factorial. So, if we use this to rewrite the numerator of our fraction, we can cancel the shared factor of 𝑛 minus three factorial in the numerator and the denominator. This gives us 𝑛 multiplied by 𝑛 minus one multiplied by 𝑛 minus two all divided by three factorial multiplied by 𝑛 cubed.

Since we have three separate factors in our numerator and a factor of 𝑛 cubed in our denominator, we’re going to divide each term in our numerator by 𝑛. This gives us 𝑛 over 𝑛 multiplied by 𝑛 minus one over 𝑛 multiplied by 𝑛 minus two over 𝑛 all divided by three factorial. We can simplify this further by noticing that 𝑛 divided by 𝑛 is just equal to one. Finally, we can divide through by 𝑛 in both terms of our numerator to get one minus one over 𝑛 multiplied by one minus two over 𝑛 all divided by three factorial. So, the fourth term in our binomial expansion of one plus one over 𝑛 all raised to the 𝑛th power was our option D.

The next part of our question wants us to find the limit of the π‘˜ plus oneth term as 𝑛 tends to ∞. To do this, we need to find an expression for the π‘˜ plus oneth term in our binomial expansion of one plus one over 𝑛 all raised to the 𝑛th power. And we do this the exact same way we did before. We substitute π‘Ÿ is equal to π‘˜, π‘₯ is equal to one, and 𝑦 is equal to one over 𝑛 into the summand in our binomial expansion formula.

This gives us that the π‘˜ plus oneth term in our expansion of one plus one over 𝑛 all raised to the 𝑛th power is 𝑛 choose π‘˜ multiplied by one to the power of 𝑛 minus π‘˜ multiplied by one over 𝑛 to the π‘˜th power. As we did previously, we rewrite 𝑛 choose π‘˜ as 𝑛 factorial divided by π‘˜ factorial multiplied by 𝑛 minus π‘˜ factorial. We have that one to the power of 𝑛 minus π‘˜ is just equal to one and one over 𝑛 all raised to the power of π‘˜ is just equal to one over 𝑛 to the π‘˜th power.

This gives us 𝑛 factorial divided by π‘˜ factorial multiplied by 𝑛 minus π‘˜ factorial multiplied by 𝑛 to the π‘˜th power. Now, we’re going to rewrite our numerator of 𝑛 factorial as 𝑛 multiplied by 𝑛 minus one. And we multiply all the way down to the term before 𝑛 minus π‘˜. And we multiply all of this by 𝑛 minus π‘˜ factorial. And we do this so we can remove the shared factor of 𝑛 minus π‘˜ factorial in our numerator and our denominator. This gives us 𝑛 multiplied by 𝑛 minus one. And we multiply all the way down to 𝑛 minus the term before π‘˜. And we divide all of this by π‘˜ factorial multiplied by 𝑛 to the π‘˜th power.

Now, we noticed that our numerator has π‘˜ separate factors and our denominator has 𝑛 to the power of π‘˜, which is π‘˜ separate factors of 𝑛. So, we can divide each of the terms in our numerator by 𝑛. This gives us the new expression of 𝑛 over 𝑛 multiplied by 𝑛 minus one over 𝑛 multiplied by all the way down to 𝑛 minus the term before π‘˜ divided by 𝑛. And we divide all of this by π‘˜ factorial. We can simplify 𝑛 divided by 𝑛 to just be equal to one.

The rest of the terms in our numerator are of the form 𝑛 minus π‘Ž all divided by 𝑛 for some constant π‘Ž. We could divide through by 𝑛 in all of these expressions to give us one minus π‘Ž divided by 𝑛. Dividing through by 𝑛 in the first term of our numerator gives us one minus one over 𝑛. Dividing through by 𝑛 in the second term in our numerator gives us one minus two over 𝑛. And this carries on all the way to the last term in our numerator. We divide through by 𝑛, which gives us one minus π‘˜ minus one over 𝑛. And we divide all of this by π‘˜ factorial.

So, we’ve now found an expression for the π‘˜ plus oneth term in our binomial expansion of one plus one over 𝑛 all raised to the 𝑛th power. And the question is asking us to calculate the limit as 𝑛 approaches ∞ of the π‘˜ plus oneth term of our expansion. One thing we can notice is that as the value of 𝑛 changes, the value of π‘˜ remains constant. So, we can take this constant value of π‘˜ outside of our limits.

Next, as our limit has 𝑛 approaching ∞, we can notice that one over 𝑛, two over 𝑛, all the way down to π‘˜ minus one over 𝑛 are all approaching zero. Since the numerator remains constant but the denominator is getting larger and larger. Now, if all of these are approaching zero, then if we look at each individual pair of parentheses, we can see that it is approaching one. So, our limit is just one multiplied by one multiplied by one, which is just equal to one. So, our limit as 𝑛 approaches ∞ of the π‘˜ plus oneth term of the binomial expansion of one plus one over 𝑛 all raised to the 𝑛th power is equal to one over π‘˜ factorial.

The next part of our question wants us to write a series in summation, or sigma, notation which is equal to the limit of one plus one over 𝑛 to the 𝑛th power as 𝑛 tends to ∞. Well, one thing we do know is the binomial expansion of one plus one over 𝑛 all raised to the 𝑛th power will be equal to the first term in its expansion plus the second term in its expansion. And we add all the way up to the last term in its expansion.

And we already found an expression for the π‘˜ plus oneth term in our binomial expansion of one plus one over 𝑛 all raised to the 𝑛th power. So, the expansion is equal to the sum from π‘˜ equals zero to 𝑛 of one minus one over 𝑛 multiplied by one minus two over 𝑛 multiplied by all the way down to one minus π‘˜ minus one over 𝑛 all divided by π‘˜ factorial. We now take the limit as 𝑛 approaches ∞ on both sides of our equation. On the right-hand side of our equation, we take the limit inside of our sum to give us the sum from π‘˜ equals zero to ∞ of the limit as 𝑛 approaches ∞ of one minus one over 𝑛 multiplied by one minus two over 𝑛 all the way down to one minus π‘˜ minus one over 𝑛. And we divide this by π‘˜ factorial.

And we see that this is now the limit as 𝑛 approaches ∞ of our π‘˜ plus oneth term, which we’ve already shown is equal to one divided by π‘˜ factorial. So, this is equal to the sum from π‘˜ equal zero to ∞ of one divided by π‘˜ factorial. And then, we can change from summing over π‘˜ to summing over 𝑛 to see that the limit as 𝑛 approaches ∞ of one plus one over 𝑛 all raised to the 𝑛th power is equal to the sum from 𝑛 equals zero to ∞ of one divided by 𝑛 factorial.

The last part of our question wants us to evaluate the value of this series. And we can do this by noticing that our series is very similar to one we’ve seen before. It’s similar to the McLaurin series for 𝑒 to the power of π‘₯. Which tells us that 𝑒 to the power of π‘₯ is equal to the sum from 𝑛 equals zero to ∞ of π‘₯ to the 𝑛th power divided by 𝑛 factorial. Now, if we substitute π‘₯ is equal to one into our McLaurin series for 𝑒 to the power of π‘₯, we have 𝑒 to the first power is equal to the sum from 𝑛 equals zero to ∞ of one to the 𝑛th power divided by 𝑛 factorial.

And since one to the 𝑛th power is just equal to one, we have that 𝑒 to the first power is equal to the sum from 𝑛 equals zero to ∞ of one divided by 𝑛 factorial. And Since 𝑒 to the first power is just equal to 𝑒, we’ve shown that our series was equal to 𝑒. Therefore, by answering this question, what we have actually shown is that the limit as 𝑛 approaches ∞ of one plus one over 𝑛 all raised to the 𝑛th power is equal to Euller’s constant 𝑒.

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