Lesson Video: General Term in the Binomial Theorem | Nagwa Lesson Video: General Term in the Binomial Theorem | Nagwa

# Lesson Video: General Term in the Binomial Theorem Mathematics

In this video, we will learn how to find a specific term inside a binomial expansion and find the relation between two consecutive terms.

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### Video Transcript

In this video, we will learn how to find a specific term inside a binomial expansion and find the relation between two consecutive terms. We will begin by looking at the binomial theorem and recall what each of its individual parts represents.

The binomial theorem provides us with a general formula for expanding binomials raised to large powers. The binomial theorem states that π plus π raised to the power of π is equal to π choose zero multiplied by π to the power of π plus π choose one multiplied by π to the power of π minus one multiplied by π to the power of one, and so on. The general term of the expansion is π choose π multiplied by π to the power of π minus π multiplied by π to the power of π. This then continues to the final term π choose π multiplied by π to the power of π. The exponent or power of π decreases, whereas the exponent of π increases.

The π choose π or π C π combinations notation stands for π factorial divided by π minus π factorial multiplied by π factorial. There are other ways of writing this as shown, but we will stick to the first notation in this video. In addition to the general theorem, weβre sometimes interested in a particular term in the expansion. This will be the focus for the first half of this video. The general term as we can see from the expansion is π C π multiplied by π to the power of π minus π multiplied by π to the power of π. The general term is notated as π subscript π plus one. This is because the first term occurs when π equals zero as we can see from our expansion. It is also important to note that there will be a total of π plus one terms in any binomial expansion.

We will now look at a question where we need to find a specific term of a binomial expansion.

Find the third term in the expansion of two π₯ plus five over the square root of π₯ to the power of five.

This is an example of a binomial expansion written in the form π plus π to the power of π. We could write out the whole expansion. However, we know that the general term π sub π plus one is equal to π choose π multiplied by π to the power of π minus π multiplied by π to the power of π. In this question, we want to find the third term π sub three. Inside the parentheses, we have two terms. π, the first term, is equal to two π₯. And π, the second term, is equal to five over root π₯.

The exponent or power that this is raised to is five. Therefore, π is equal to five. As we are trying to find the third term, π plus one is equal to three. Subtracting one from both sides of this equation gives us π is equal to two. We can now substitute these four values into the general term formula. The third term is therefore equal to five choose two multiplied by two π₯ cubed multiplied by five over the square root of π₯ squared.

We know that π choose π is equal to π factorial divided by π minus π factorial multiplied by π factorial. Five choose two is therefore equal to five factorial divided by three factorial multiplied by two factorial. We can rewrite five factorial as five multiplied by four multiplied by three factorial. This simplifies to five multiplied by four divided by two factorial, which equals 10. Five choose two is equal to 10.

When cubing two π₯, we can cube the two and π₯ separately. As two cubed is equal to eight, two π₯ cubed is equal to eight π₯ cubed. We can use a similar method when squaring a fraction. We square the numerator and denominator separately. Five squared is equal to 25, and the square root of π₯ squared is π₯. A square rooting and squaring are inverse operations. The third term is therefore equal to 10 multiplied by eight π₯ cubed multiplied by 25 over π₯.

This can be simplified to 2000π₯ squared as 25 multiplied by eight multiplied by 10 is 2000 and π₯ cubed divided by π₯ is π₯ squared. The third term in the expansion of two π₯ plus five over the square root of π₯ to the power of five is 2000π₯ squared. To save time, we couldβve also calculated five choose two on the calculator.

In our next question, we will look at two different terms in a binomial expansion.

Consider the expansion of π₯ to the fifth power over eight minus eight over π₯ all raised to the ninth power in descending powers of π₯. For which values of π₯ is the sum of the middle two terms equal to zero?

In this question, we have a binomial expression written in the form π plus π raised to the power of π. When expanding any expression of this type, we know it will have π plus one terms. This means that our expansion will have 10 terms, and the middle two are the fifth and the sixth. Weβre interested in π sub five and π sub six, the fifth and sixth terms of the expansion.

We know the general term of any binomial expansion π sub π plus one is equal to π choose π multiplied by π to the power of π minus π multiplied but π to the power of π. The fifth term is therefore equal to nine choose four multiplied by π₯ to the fifth power over eight to the fifth power multiplied by negative eight over π₯ to the fourth power.

The sixth term, on the other hand, is equal to nine choose five multiplied by π₯ to the fifth power over eight to the fourth power multiplied by negative eight over π₯ to the fifth power. We are told that the sum of these two terms is equal to zero. This means that the fifth term is equal to the negative of the sixth term. We notice that nine choose four is equal to nine choose five as they are both equal to nine factorial divided by five factorial multiplied by four factorial.

We can therefore cancel this on both sides of our equation. Both sides of the equation also can be divided by π₯ to the fifth power over eight raised to the fourth power. This means that the left-hand side becomes π₯ to the fifth power over eight. We can also divide both sides by negative eight over π₯ to the fourth power. The right-hand side becomes negative negative eight over π₯. The two negatives become a positive.

We can then cross multiply. We can multiply both sides by eight and π₯. This gives us π₯ to the sixth power is equal to 64. We can then take the sixth root of both sides. This gives us π₯ is equal to positive or negative two as positive and negative two both raised to the power of six give us 64.

The word βvaluesβ in the question suggests we will have more than one answer. The sum of the middle two terms is equal to zero when π₯ is equal to negative two or two.

In the remainder of this video, we will consider what happens when we look at the ratio between consecutive terms of a binomial expansion.

Consider the expansion of eight π₯ plus two π¦ to the 23rd power. Find the ratio between the eighth and seventh terms.

We recall that the general term π sub π plus one is equal to π choose π multiplied by π to the power of π minus π multiplied but π to the power of π. This means that the ratio between the eighth and seventh terms, π eight divided by π seven, is equal to 23 choose seven multiplied by eight π₯ to the power of 16 multiplied by two π¦ to the power of seven divided by 23 choose six multiplied by eight π₯ to the power of 17 multiplied by two π¦ to the power of six.

We immediately see that we can divide the numerator and denominator by eight π₯ to the power of 16. This leaves us with eight π₯ on the numerator. We note that this was equal to the first term in our binomial expansion.

We can also divide the numerator and denominator by two π¦ to the power of six. This leaves us with two π¦ on the numerator, which was equal to the second term in our expansion. We recall that the ratio of consecutive combinations π choose π divided by π choose π minus one is given by π minus π plus one over π. This means that 23 choose seven divided by 23 choose six is equal to 23 minus seven plus one all over seven. This simplifies to 17 over seven.

The ratio of the eighth term divided by the seventh term is therefore equal to 17 multiplied by two π¦ divided by seven multiplied by eight π₯. This in turn simplifies to 17π¦ over 28π₯. This is the ratio between the eighth and seventh terms.

This example leads us to a simple expression for the general ratio between consecutive terms. If we have two consecutive terms, π sub π plus one and π sub π, of the expansion π plus π to the πth power, then the ratio of these two terms is equal to π choose π multiplied by π to the power of π minus π multiplied by π to the power of π all divided by π choose π minus one multiplied by π to the power of π minus π plus one multiplied by π to the power of π minus one. This simplifies to π minus π plus one over π multiplied by π over π. We can quote this formula to help us solve problems involving the ratios of consecutive terms in binomial expansions.

In our final question, we will look at the ratio between nonconsecutive terms.

Find the ratio between the 15th and 17th terms in the expansion of π₯ minus 12 to the 19th power.

In order to answer this question, we will use two formulas linked to the expansion of a binomial expression of the form π plus π to the πth power. We know that the general term of this expansion, π sub π plus one, is equal to π choose π multiplied by π to the power of π minus π multiplied by π to the power of π. We also know that the ratio of consecutive terms of a binomial expansion, π sub π plus one over π sub π, is equal to π minus π plus one over π multiplied by π over π.

In this question, we are dealing with the 15th and 17th terms in the expansion. The 15th term is equal to 19 choose 14 multiplied by π₯ to the fifth power multiplied by negative 12 to the 14th power. The 17th term is equal to 19 choose 16 multiplied by π₯ cubed multiplied by negative 12 to the 16th power. We can divide π₯ to the fifth power and π₯ cubed by π₯ cubed, leaving us with π₯ squared on the numerator. Likewise, dividing the numerator and denominator by negative 12 to the 14th power leaves us with negative 12 squared on the denominator. This is the same as π squared over π squared.

We now need to consider what happens when we divide nonconsecutive combinations. As the 17th term is two terms after the 15th term, we have π sub π over π sub π plus two. The combinations part of this will be equal to π choose π minus one over π choose π plus one. When writing this in terms of factorials, it looks quite complicated. However, we will quite quickly see that some terms cancel. We can then use our properties of factorials and the knowledge that dividing a fraction by another fraction is the same as multiplying the first fraction by the reciprocal of the second. The combinations part is therefore equal to π multiplied by π plus one divided by π minus π plus one multiplied by π minus π.

We can now see the link between this and the ratio of consecutive terms. We can now go back to our question to calculate 19 choose 14 divided by 19 choose 16. As π is equal to 19 and π is equal to 15, we have 15 multiplied by 16 divided by five multiplied by four. We know that negative 12 squared is 144. So we need to multiply the first part by π₯ squared over 144. This in turn simplifies to 12π₯ squared over 144. Finally, we can divide the numerator and denominator by 12 so that the ratio between the 15th and 17th terms is π₯ squared over 12.

We will now summarize the key points from this video. The general term in the binomial expansion of π plus π to the πth power is denoted by π sub π plus one. This is equal to π choose π multiplied by π to the power of π minus π multiplied by π to the power of π. Consecutive terms in a binomial expansion are related by the formula π sub π plus one over π sub π is equal to π minus π plus one over π multiplied by π over π. We also saw in our last question that this formula can be manipulated when dealing with ratios of nonconsecutive terms.