Explainer: General Term in the Binomial Theorem

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

The binomial theorem provides us with a general formula for expanding binomials raised to arbitrarily large powers. Being confident at using this proves extremely useful for more advanced topics in mathematics. We begin by recalling the statement of the binomial theorem.

Binomial Theorem

For an integer ๐‘›, (๐‘Ž+๐‘)=๐ถ๐‘Ž+๐ถ๐‘Ž๐‘+๐ถ๐‘Ž๐‘+โ‹ฏ+๐ถ๐‘Ž๐‘+โ‹ฏ+๐ถ๐‘Ž๐‘+๐ถ๐‘,๏Š๏Š๏Šฆ๏Š๏Š๏Šง๏Š๏Šฑ๏Šง๏Šง๏Š๏Šจ๏Š๏Šฑ๏Šจ๏Šจ๏Š๏Ž๏Š๏Šฑ๏Ž๏Ž๏Š๏Š๏Šฑ๏Šง๏Šง๏Š๏Šฑ๏Šง๏Š๏Š๏Š where ๏Š๏Ž๐ถ=๐‘›!(๐‘›โˆ’๐‘Ÿ)!๐‘Ÿ!.

Sometimes the following notations are used in place of ๏Š๏Ž๐ถ: ๏€ฟ๐‘›๐‘Ÿ๏‹, ๏Š๏Ž๐ถ, ๐ถ๏Š๏Ž, ๐ถ๏Š๏Ž•๏Ž, and ๐ถ(๐‘›,๐‘Ÿ).

In addition to using the general theorem, we are sometimes interested in considering a particular term in the expansion. For this, we use the formula for the general term presented below.

General Term of the Binomial Expansion

In the expansion of (๐‘Ž+๐‘)๏Š, the general term is given by ๏Š๏Ž๏Š๏Šฑ๏Ž๏Ž๐ถ๐‘Ž๐‘.

The general term is variously notated ๐‘‡๏Ž๏Šฐ๏Šง or ๐‘Ž๏Ž๏Šฐ๏Šง. The important thing to note when referring to terms by order is that the first term is the term for which ๐‘Ÿ=0. Hence, ๐‘‡=๐‘Ž=๐ถ๐‘Ž๐‘.๏Ž๏Šฐ๏Šง๏Ž๏Šฐ๏Šง๏Š๏Ž๏Š๏Šฑ๏Ž๏Ž

This explainer will focus on using the general term to solve problems where we are interested in particular terms in a binomial expansion. In many of these questions, we can always resort to fully expanding the binomial. However, this is often laborious, and using the general term leads to simpler solutions which are less prone to error.

In general, we need to be careful to remember that the first term of a binomial expansion is the term for which ๐‘Ÿ=0. It is a common mistake to assume that the first term is when ๐‘Ÿ=1. However, this is incorrect and hence why we tend to define ๐‘Ž๏Ž๏Šฐ๏Šง and not ๐‘Ž๏Ž to reinforce this fact. Even though we could write the terms of a binomial expansion in any order, there is a standard order which is assumed in most questions which ask for the second, third, or tenth term. The standard order for the terms of the expansion of (๐‘Ž+๐‘)๏Š is decreasing powers of ๐‘Ž and increasing powers of ๐‘.

Example 1: Finding the ๐‘›th Term in a Binomial Expansion

Find the third term in the expansion of ๏€ฟ2๐‘ฅ+5โˆš๐‘ฅ๏‹๏Šซ.

Answer

When presented with a question like this, it would be perfectly legitimate to write the full expansion and then take the coefficient from the appropriate term. However, appealing to the formula for the general term simplifies our calculation. This is the method we will demonstrate here. Recall that the formula for the general term for the expansion of (๐‘Ž+๐‘)๏Š is ๏Š๏Ž๏Š๏Šฑ๏Ž๏Ž๐ถ๐‘Ž๐‘.

However, we need to remember that the first term starts from ๐‘Ÿ=0. Hence, the third term will be given by ๐‘Ÿ=2 and not, as we might naรฏvely think, by ๐‘Ÿ=3. Hence, by setting ๐‘Ÿ=2, ๐‘Ž=2๐‘ฅ, ๐‘=5โˆš๐‘ฅ, and ๐‘›=5, we have ๏Šซ๏Šจ๏Šฉ๏Šจ๏Šฉ๏Šฉ๏Šจ๐ถ(2๐‘ฅ)๏€ฟ5โˆš๐‘ฅ๏‹=10ร—2๐‘ฅ25๐‘ฅ=2,000๐‘ฅ.

Notice that by using the general term, we can often simplify the calculations we need to make.

Example 2: Finding a Given Term in a Binomial Expansion

Find ๐‘Ž๏Šช in the expansion of ๏€ฟ5โˆš๐‘ฅ+โˆš๐‘ฅ5๏‹๏Šฏ.

Answer

For the expansion of (๐‘Ž+๐‘)๏Š, the general term ๐‘Ž๏Ž is defined as follows: ๐‘Ž=๐ถ๐‘Ž๐‘.๏Ž๏Šฐ๏Šง๏Š๏Ž๏Š๏Šฑ๏Ž๏Ž

Hence, by setting ๐‘›=9, ๐‘Ÿ=3, ๐‘Ž=5โˆš๐‘ฅ, and ๐‘=โˆš๐‘ฅ5, we have ๐‘Ž=๐ถ๏€ฟ5โˆš๐‘ฅ๏‹๏€ฟโˆš๐‘ฅ5๏‹.๏Šช๏Šฏ๏Šฉ๏Šฌ๏Šฉ

Since 5โˆš๐‘ฅ=๏€ฟโˆš๐‘ฅ5๏‹๏Šฑ๏Šง, we can rewrite this as ๐‘Ž=๐ถ๏€ฟ5โˆš๐‘ฅ๏‹๏€ฟ5โˆš๐‘ฅ๏‹=84๏€ฟ5โˆš๐‘ฅ๏‹=10,500๐‘ฅ.๏Šช๏Šฏ๏Šฉ๏Šฌ๏Šฑ๏Šฉ๏Šฉ๏Šฑ๏Žข๏Žก

In the following example, we will see how, by using the general term, we are able to solve for unknowns.

Example 3: Using the General Term to Find Unknowns

The terms of the expansion of (2๐‘ฅ+๐‘š๐‘ฆ)๏Šซ are arranged according to the descending powers of ๐‘ฅ. Given that ๐‘Ž=2,560๐‘ฅ๐‘ฆ๏Šช๏Šจ๏Šฉ, find the value of ๐‘š.

Answer

To solve this problem, we can use the formula for the general term of the binomial expansion to find an alternative expression for ๐‘Ž๏Šช. We can then equate the two expressions and solve for ๐‘š. Recall that the general term of the binomial expansion of (๐‘+๐‘ž)๏Š is given by ๐‘Ž=๐ถ๐‘๐‘ž.๏Ž๏Šฐ๏Šง๏Š๏Ž๏Š๏Šฑ๏Ž๏Ž

Setting ๐‘=2๐‘ฅ, ๐‘ž=๐‘š๐‘ฆ, ๐‘›=5, and ๐‘Ÿ=3, we have ๐‘Ž=๐ถ(2๐‘ฅ)(๐‘š๐‘ฆ)=10ร—2๐‘š๐‘ฅ๐‘ฆ=40๐‘š๐‘ฅ๐‘ฆ.๏Šช๏Šซ๏Šฉ๏Šจ๏Šฉ๏Šจ๏Šฉ๏Šจ๏Šฉ๏Šฉ๏Šจ๏Šฉ

In the question, we are told that ๐‘Ž=2,560๐‘ฅ๐‘ฆ๏Šช๏Šจ๏Šฉ. Hence, we can equate these two expressions for ๐‘Ž๏Šช and solve for ๐‘š as follows: 2,560๐‘ฅ๐‘ฆ=40๐‘š๐‘ฅ๐‘ฆ.๏Šจ๏Šฉ๏Šฉ๏Šจ๏Šฉ

Cancelling the common factors, we have 64=๐‘š.๏Šฉ

Hence, ๐‘š=4.

Example 4: Using the General Term

If the coefficient of the third term in the expansion of ๏€ผ๐‘ฅโˆ’14๏ˆ๏Š is 338, determine the middle term in the expansion.

Answer

Using the formula for the general term of the binomial expansion, we can find an expression for the coefficient of the third term in terms of ๐‘›. Using this, we can solve for ๐‘› and then find the middle term of the expansion. Recall that the general term of the binomial expansion of (๐‘Ž+๐‘)๏Š is given by ๐‘Ž=๐ถ๐‘Ž๐‘.๏Ž๏Šฐ๏Šง๏Š๏Ž๏Š๏Šฑ๏Ž๏Ž

Setting ๐‘Ž=๐‘ฅ, ๐‘=14, and ๐‘Ÿ=2, we have ๐‘Ž=๐ถ๐‘ฅ๏€ผ14๏ˆ.๏Šฉ๏Š๏Šจ๏Š๏Šฑ๏Šจ๏Šจ

Using the definition of ๏Š๏Ž๐ถ, we can rewrite this as ๐‘Ž=๐‘›(๐‘›โˆ’1)2๐‘ฅร—116.๏Šฉ๏Š๏Šฑ๏Šจ

Since we know the coefficient of this term is 338, we can write ๐‘›(๐‘›โˆ’1)32=338.

Multiplying both sides of the equation by 32, we have ๐‘›โˆ’๐‘›=132.๏Šจ

Hence, we can find ๐‘› by solving the quadratic equation ๐‘›โˆ’๐‘›โˆ’132=0.๏Šจ

We can solve this by factoring or using the quadratic formula to find (๐‘›โˆ’12)(๐‘›+11)=0.

Hence, ๐‘›=12 or ๐‘›=โˆ’11. Since ๐‘› is a positive integer, we can discard the negative solutions and, hence, ๐‘›=12. We can now use this to find the middle term of the expansion. Since ๐‘›=12, there will be thirteen terms in the expansion, and the middle term will be the seventh term. Hence, we can use the formula for the general term to find ๐‘Ž๏Šญ as follows: ๐‘Ž=๐ถ๐‘ฅ๏€ผ14๏ˆ=9244,096๐‘ฅ=2311,024๐‘ฅ.๏Šญ๏Šง๏Šจ๏Šฌ๏Šง๏Šจ๏Šฑ๏Šฌ๏Šฌ๏Šฌ๏Šฌ

Example 5: Finding the Ratio between Consecutive Terms

Consider the expansion of (8๐‘ฅ+2๐‘ฆ)๏Šจ๏Šฉ. Find the ratio between the eighth and the seventh terms.

Answer

Recall that the formula for the general term of the binomial expansion of (๐‘Ž+๐‘)๏Š is ๐‘Ž=๐ถ๐‘Ž๐‘.๏Ž๏Šฐ๏Šง๏Š๏Ž๏Š๏Šฑ๏Ž๏Ž

Therefore, we can write the general term of the expansion of (8๐‘ฅ+2๐‘ฆ)๏Šจ๏Šฉ by setting ๐‘Ž=8๐‘ฅ, ๐‘ฆ=2๐‘ฆ, and ๐‘›=23 as follows: ๐‘Ž=๐ถ(8๐‘ฅ)(2๐‘ฆ)=๐ถ82๐‘ฅ๐‘ฆ.๏Ž๏Šฐ๏Šง๏Šจ๏Šฉ๏Ž๏Šจ๏Šฉ๏Šฑ๏Ž๏Ž๏Šจ๏Šฉ๏Ž๏Šจ๏Šฉ๏Šฑ๏Ž๏Ž๏Šจ๏Šฉ๏Šฑ๏Ž๏Ž

Therefore, the ratio between the eighth and seventh term is given by ๐‘Ž๐‘Ž=๐ถ82๐‘ฅ๐‘ฆ๐ถ82๐‘ฅ๐‘ฆ.๏Šฎ๏Šญ๏Šจ๏Šฉ๏Šญ๏Šจ๏Šฉ๏Šฑ๏Šญ๏Šญ๏Šจ๏Šฉ๏Šฑ๏Šญ๏Šญ๏Šจ๏Šฉ๏Šฌ๏Šจ๏Šฉ๏Šฑ๏Šฌ๏Šฌ๏Šจ๏Šฉ๏Šฑ๏Šฌ๏Šฌ

Using the rules of exponents, we can simplify this to ๐‘Ž๐‘Ž=๐ถ2๐‘ฆ๐ถ82๐‘ฅ=๐ถ๐‘ฆ๐ถ4๐‘ฅ.๏Šฎ๏Šญ๏Šจ๏Šฉ๏Šญ๏Šจ๏Šฉ๏Šฌ๏Šฌ๏Šจ๏Šฉ๏Šญ๏Šจ๏Šฉ๏Šฌ

Recall that the ratio of consecutive combinations is given by ๏Š๏Ž๏Š๏Ž๏Šฑ๏Šง๐ถ๐ถ=๐‘›โˆ’๐‘Ÿ+1๐‘Ÿ.

Hence, ๏Šจ๏Šฉ๏Šญ๏Šจ๏Šฉ๏Šฌ๐ถ๐ถ=23โˆ’7+17=177.

Substituting this into the equation above, we have ๐‘Ž๐‘Ž=17๐‘ฆ7ร—4๐‘ฅ=17๐‘ฆ28๐‘ฅ.๏Šฎ๏Šญ

In the last example, we considered the ratio between two consecutive terms. This is in fact a common thing to consider and there is a simple expression for this in general. Consider the two consecutive terms ๐‘Ž๏Ž๏Šฐ๏Šง and ๐‘Ž๏Ž of the expansion of (๐‘Ž=๐‘)๏Š; using the formula for the general term, we can write their ratio as follows: ๐‘Ž๐‘Ž=๐ถ๐‘Ž๐‘๐ถ๐‘Ž๐‘.๏Ž๏Šฐ๏Šง๏Ž๏Š๏Ž๏Š๏Šฑ๏Ž๏Ž๏Š๏Ž๏Šฑ๏Šง๏Š๏Šฑ๏Ž๏Šฐ๏Šง๏Ž๏Šฑ๏Šง

Using the rules of exponents, we can simplify this to ๐‘Ž๐‘Ž=๐ถ๐‘๐ถ๐‘Ž.๏Ž๏Šฐ๏Šง๏Ž๏Š๏Ž๏Š๏Ž๏Šฑ๏Šง

We can now use the formula for ratios of consecutive combinations: ๏Š๏Ž๏Š๏Ž๏Šฑ๏Šง๐ถ๐ถ=๐‘›โˆ’๐‘Ÿ+1๐‘Ÿ, to rewrite this as ๐‘Ž๐‘Ž=(๐‘›โˆ’๐‘Ÿ+1)๐‘Ÿ๐‘๐‘Ž.๏Ž๏Šฐ๏Šง๏Ž

We can use this formula to help us solve problems involving the ratios of consecutive terms in binomial expansions.

Example 6: Using Ratios between Consecutive Terms to Solve for Unknowns

Consider the expansion of (๐‘Ž+๐‘)๏Š. Find the values of ๐‘Ž, ๐‘, and ๐‘› given that ๐‘Ž=215,040๏Šซ, ๐‘Ž=258,048๏Šฌ, and ๐‘Ž=215,040๏Šญ.

Answer

One of the simplest ways to address this problem is by considering the ratios of consecutive terms. Recall that the ratio of two consecutive terms ๐‘Ž๏Ž๏Šฐ๏Šง and ๐‘Ž๏Ž in the expansion of (๐‘Ž+๐‘)๏Š is given by ๐‘Ž๐‘Ž=(๐‘›โˆ’๐‘Ÿ+1)๐‘Ÿ๐‘๐‘Ž.๏Ž๏Šฐ๏Šง๏Ž

Hence, we have

๐‘Ž๐‘Ž=(๐‘›โˆ’5)6๐‘๐‘Ž,๐‘Ž๐‘Ž=(๐‘›โˆ’4)5๐‘๐‘Ž.๏Šญ๏Šฌ๏Šฌ๏Šซ(1)

By considering the ratio of these two terms, we can eliminate ๐‘Ž and ๐‘ and be left with an equation in terms of ๐‘›. Hence, ๏Œบ๏Œบ๏Œบ๏Œบ(๏Š๏Šฑ๏Šซ)๏Šฌ๏Œป๏Œบ(๏Š๏Šฑ๏Šช)๏Šซ๏Œป๏Œบ๏Žฆ๏Žฅ๏Žฅ๏Žค=.

Simplifying this, we get ๐‘Ž๐‘Ž๐‘Ž=5(๐‘›โˆ’5)6(๐‘›โˆ’4).๏Šญ๏Šซ๏Šจ๏Šฌ

Substituting in the values of ๐‘Ž๏Šซ, ๐‘Ž๏Šฌ, and ๐‘Ž๏Šญ, we have 56=5(๐‘›โˆ’5)6(๐‘›โˆ’4).๏Šจ๏Šจ

Dividing both sides by 56 gives 56=๐‘›โˆ’5๐‘›โˆ’4.

Cross multiplying by (๐‘›โˆ’4) and 6, we have 5(๐‘›โˆ’4)=6(๐‘›โˆ’5).

Hence, ๐‘›=10.

By substituting our value of ๐‘› into equation (1), we have ๐‘Ž๐‘Ž=56๐‘๐‘Ž.๏Šญ๏Šฌ

Using the values of ๐‘Ž๏Šฌ and ๐‘Ž๏Šญ, we find 56=56๐‘๐‘Ž.

Hence, ๐‘Ž=๐‘. We can now use the formula for the general term to find the value of ๐‘Ž and ๐‘ as follows. Using the term ๐‘Ž๏Šซ, we have 215,040=๐ถ๐‘Ž๐‘.๏Šง๏Šฆ๏Šช๏Šง๏Šฆ๏Šฑ๏Ž๏Ž

Since ๐‘Ž=๐‘, we can rewrite this as 215,040=๐ถ๐‘Ž=210๐‘Ž,๏Šง๏Šฆ๏Šช๏Šง๏Šฆ๏Šง๏Šฆ which simplifies to 1,024=๐‘Ž.๏Šง๏Šฆ

Hence, ๐‘Ž=2. Therefore, the final answer is ๐‘Ž=๐‘=2 and ๐‘›=10.

Example 7: Using the Ratio of Consecutive Terms

Consider the binomial expansion of (3+7๐‘ฅ)๏Šจ๏Šฎ in ascending powers of ๐‘ฅ. When ๐‘ฅ=6, one of the terms in the expansion is equal to twice its following term. Find the position of these two terms.

Answer

The question states that when ๐‘ฅ=6, one of the terms in the expansion is equal to twice its following term. We can write this algebraically as ๐‘Ž=2๐‘Ž.๏Ž๏Ž๏Šฐ๏Šง

Hence, ๐‘Ž๐‘Ž=12.๏Ž๏Šฐ๏Šง๏Ž

Recall that for the binomial expansion of (๐‘Ž+๐‘)๏Š, the ratio between consecutive terms is given by ๐‘Ž๐‘Ž=(๐‘›โˆ’๐‘Ÿ+1)๐‘๐‘Ÿ๐‘Ž.๏Ž๏Šฐ๏Šง๏Ž

Setting ๐‘Ž=3, ๐‘=7๐‘ฅ, and ๐‘›=28, we can rewrite this as ๐‘Ž๐‘Ž=(28โˆ’๐‘Ÿ+1)7๐‘ฅ3๐‘Ÿ.๏Ž๏Šฐ๏Šง๏Ž

Since this is equal to a half when ๐‘ฅ=6, we can write 12=(29โˆ’๐‘Ÿ)7ร—63๐‘Ÿ.

Multiplying through by 2๐‘Ÿ gives ๐‘Ÿ=28(29โˆ’๐‘Ÿ)=812โˆ’28๐‘Ÿ.

Gathering like terms, we have 29๐‘Ÿ=812.

Hence, ๐‘Ÿ=28. Therefore, the two terms that satisfy the given condition are ๐‘Ž๏Šจ๏Šฎ and ๐‘Ž๏Šจ๏Šฏ.

Example 8: Ratios of Nonconsecutive Terms

Consider the expansion of (๐‘š๐‘ฅ+8)๏Š. Determine the values of ๐‘š and ๐‘›, given that the ratio between the coefficients of ๐‘Ž๏Šง๏Šจ and ๐‘Ž๏Šง๏Šช is equal to 6374,640 and that the ratio between the coefficients of ๐‘Ž๏Šญ and ๐‘Ž๏Šฏ is equal to 491,360.

Answer

We would now like to derive an algebraic expression for the ratio of the two terms ๐‘Ž๏Ž and ๐‘Ž๏Ž๏Šฐ๏Šจ. We can do this using the formula for the general term of a binomial expansion. For the expansion of (๐‘š๐‘ฅ+8)๏Š, we can write the general term as ๐‘Ž=๐ถ(๐‘š๐‘ฅ)8.๏Ž๏Š๏Ž๏Šฑ๏Šง๏Š๏Šฑ๏Ž๏Šฐ๏Šง๏Ž๏Šฑ๏Šง

Therefore, we can write the ratio of the two terms ๐‘Ž๏Ž and ๐‘Ž๏Ž๏Šฐ๏Šจ as follows: ๐‘Ž๐‘Ž=๐ถ(๐‘š๐‘ฅ)8๐ถ(๐‘š๐‘ฅ)8.๏Ž๏Ž๏Šฐ๏Šจ๏Š๏Ž๏Šฑ๏Šง๏Š๏Šฑ๏Ž๏Šฐ๏Šง๏Ž๏Šฑ๏Šง๏Š๏Ž๏Šฐ๏Šง๏Š๏Šฑ๏Ž๏Šฑ๏Šง๏Ž๏Šฐ๏Šง

Using the powers of exponents, we can rewrite this as ๐‘Ž๐‘Ž=๐ถ๐‘š๐‘ฅ๐ถ8.๏Ž๏Ž๏Šฐ๏Šจ๏Š๏Ž๏Šฑ๏Šง๏Šจ๏Šจ๏Š๏Ž๏Šฐ๏Šง๏Šจ

Therefore, the ratios of the coefficients of these terms, which we will denote ๐‘…๏Ž, is given by ๐‘…=๐ถ๐‘š๐ถ8.๏Ž๏Š๏Ž๏Šฑ๏Šง๏Šจ๏Š๏Ž๏Šฐ๏Šง๏Šจ

Using the definition of ๏Š๏Ž๐ถ, we can further simplify this expression as follows: ๐‘…=๐‘š8=(๐‘›โˆ’๐‘Ÿโˆ’1)!(๐‘Ÿ+1)!(๐‘›โˆ’๐‘Ÿ+1)!(๐‘Ÿโˆ’1)!๐‘š8.๏Ž๏Š!(๏Š๏Šฑ๏Ž๏Šฐ๏Šง)!(๏Ž๏Šฑ๏Šง)!๏Š!(๏Š๏Šฑ๏Ž๏Šฑ๏Šง)!(๏Ž๏Šฐ๏Šง)!๏Šจ๏Šจ๏Šจ๏Šจ

Using the properties of factorials, we have ๐‘…=๐‘Ÿ(๐‘Ÿ+1)(๐‘›โˆ’๐‘Ÿ+1)(๐‘›โˆ’๐‘Ÿ)๐‘š8.๏Ž๏Šจ๏Šจ

In the question, we are given two facts that we can express algebraically as ๐‘…=6374,640๏Šง๏Šจ and ๐‘…=491,360.๏Šญ

To solve for ๐‘›, we can take the ratio of ๐‘…๏Šง๏Šจ to ๐‘…๏Šญ; this will result in eliminating ๐‘š from the equation, and we will simply be left with an equation in ๐‘› which we will be able to solve: ๐‘…๐‘…==156(๐‘›โˆ’6)(๐‘›โˆ’7)56(๐‘›โˆ’11)(๐‘›โˆ’12)=3914(๐‘›โˆ’6)(๐‘›โˆ’7)(๐‘›โˆ’11)(๐‘›โˆ’12).๏Šง๏Šจ๏Šญ๏Šง๏Šจ(๏Šง๏Šฉ)(๏Š๏Šฑ๏Šง๏Šง)(๏Š๏Šฑ๏Šง๏Šจ)๏Šญ(๏Šฎ)(๏Š๏Šฑ๏Šฌ)(๏Š๏Šฑ๏Šญ)

Using the values of ๐‘…๏Šญ and ๐‘…๏Šง๏Šจ from the question, we have ๐‘…๐‘…=6374,640ร—1,36049=22158.๏Šง๏Šจ๏Šญ

Equating these two equations, we have 22158=3914(๐‘›โˆ’6)(๐‘›โˆ’7)(๐‘›โˆ’11)(๐‘›โˆ’12).

Multiplying by (๐‘›โˆ’11)(๐‘›โˆ’12) and dividing by 3914 yield 11987(๐‘›โˆ’11)(๐‘›โˆ’12)=(๐‘›โˆ’6)(๐‘›โˆ’7).

Multiplying by 87, and expanding the parentheses, we have 119๏€น๐‘›โˆ’23๐‘›+132๏…=87๏€น๐‘›โˆ’13๐‘›+42๏….๏Šจ๏Šจ

Gathering all the terms on the left-hand side, we have 32๐‘›โˆ’1,606๐‘›+12,054=0.๏Šจ

We can solve this equation using the quadratic formula to find solutions of ๐‘›=41 and ๐‘›=14716. Since ๐‘› must be a positive integer, we can discard the fractional solution and conclude that ๐‘›=41. Finally, substituting ๐‘›=41 into the formula for ๐‘…๏Šญ, we can solve for ๐‘š: ๐‘…=56(41โˆ’6)(41โˆ’7)๐‘š64=561,190ร—64๐‘š=11,360๐‘š.๏Šญ๏Šจ๏Šจ๏Šจ

Since ๐‘…=491,360๏Šญ, we have 11,360๐‘š=491,360.๏Šจ

Hence, ๐‘š=49,๏Šจ which gives ๐‘š=ยฑ7.

Key Points

  1. Using the general term for the binomial expansion often simplifies calculations in which we are only interested in specific terms.
  2. Consecutive terms in a binomial expansion are related by the formula ๐‘Ž๐‘Ž=(๐‘›โˆ’๐‘Ÿ+1)๐‘Ÿ๐‘๐‘Ž.๏Ž๏Šฐ๏Šง๏Ž The relationship between consecutive terms can be useful if we only require the first few terms of an expansion.

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