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

Lesson Explainer: General Term in the Binomial Theorem Mathematics

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 the binomial theorem proves extremely useful for more advanced topics in mathematics. We begin by recalling the statement of the binomial theorem.

Theorem: Binomial Theorem

For an integer 𝑛, (𝑝+𝑞)=𝐶𝑝+𝐶𝑝𝑞+𝐶𝑞𝑞++𝐶𝑝𝑞++𝐶𝑝𝑞+𝐶𝑞, where 𝐶=𝑛𝑛𝑟𝑟.

It is worth noting that if you read more widely into this topic, you may come across alternative notations for 𝐶, namely 𝑛𝑟, 𝐶, 𝐶, and 𝐶(𝑛,𝑟).

In addition to using the general theorem, we can consider a particular term in the expansion. For this, we use the formula for the general term presented below.

Formula: General Term of the Binomial Expansion

In the expansion of (𝑝+𝑞), the general term (𝑇) is 𝑇=𝐶𝑝𝑞𝑟=0,1,,𝑛.for

The important thing to note here, when referring to terms by their order, is that the first term, 𝑇, is the term for which 𝑟=0.

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

As previously mentioned, we need to remember that the first term of a binomial expansion is the term for which 𝑟=0. It is a common mistake to assume that the first term is when 𝑟=1. However, this is incorrect and hence why we tend to define 𝑇 rather than 𝑇 to reinforce this fact. Even though we could write the terms of a binomial expansion in any order, there is a standard order that is assumed in most questions that ask for the second, third, or perhaps tenth term. The standard order for the terms of the expansion of (𝑝+𝑞) is decreasing powers of 𝑝 and increasing powers of 𝑞.

Example 1: Finding a Specific Term in a Binomial Expansion

Find the third term in the expansion of 2𝑥+5𝑥.

Answer

When presented with a question like this, it would be perfectly legitimate to write the full expansion and then identify the third 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 (𝑝+𝑞) is 𝑇=𝐶𝑝𝑞.

Recall that the first term in the expansion corresponds to the general term with 𝑟=0. Hence, the third term will be given by 𝑟=2 rather than 𝑟=3. Hence, by setting 𝑟=2, 𝑝=2𝑥, 𝑞=5𝑥, and 𝑛=5, we have 𝐶(2𝑥)5𝑥=10×2𝑥×25𝑥=2000𝑥.

Therefore, the third term of the expansion is 2000𝑥.

Notice that, by using the general term, we can often simplify the calculations we need to make. In our second example, we will look at a very similar concept but with a binomial raised to a higher power.

Example 2: Finding a Given Term in a Binomial Expansion

Find 𝑇 in the expansion of 5𝑥+𝑥5.

Answer

For the expansion of (𝑝+𝑞), the general term 𝑇 is defined as follows: 𝑇=𝐶𝑝𝑞𝑟=0,1,,𝑛.for

Hence, by setting 𝑛=9, 𝑟=3, 𝑝=5𝑥, and 𝑞=𝑥5, we have 𝑇=𝐶5𝑥𝑥5.

Since 5𝑥=𝑥5, we can rewrite this as 𝑇=𝐶5𝑥5𝑥=845𝑥.

We can then simplify this using the laws of exponents as follows: =845𝑥=10500𝑥=10500𝑥.

Therefore, the fourth term of the expansion, 𝑇, is equal to 10500𝑥.

As we have seen in the previous two examples, we can use the general term of a binomial expansion to find a specified term of the expansion, but equally, we could be asked to identify the coefficient of a specified term as we will demonstrate in our next example.

Example 3: Finding the Coefficient of a Specified Term in a Binomial Expansion

Determine the coefficient of 𝑥 in the expansion of 𝑥+1𝑥.

Answer

The first thing to note in this example is that we can rewrite the binomial expression as 𝑥+𝑥. We then need to identify any terms that result in an exponent of 6. Recall that the general term of the expansion of (𝑝+𝑞) is 𝐶𝑝𝑞. We have 𝑝=𝑥, 𝑞=𝑥, and 𝑛=6, so substituting these expressions into the general term leads to 𝐶𝑥𝑥=𝐶𝑥𝑥.

Using the power law, we can write this as 𝐶𝑥=𝐶𝑥.

Since we want the term with 𝑥, we want the exponent of 𝑥 to be equal to 6. This means 63𝑟=6, which leads to 𝑟=4. We can verify this by substituting 𝑟=4 into the general term: 𝐶𝑥𝑥=𝐶𝑥×𝑥=𝐶𝑥.

We can see that the coefficient of the term is exactly 𝐶=15. Hence, the coefficient of 𝑥 is 15.

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

Example 4: Using the General Term to Find Unknowns

The terms of the expansion of (2𝑥+𝑚𝑦) are arranged according to the descending powers of 𝑥. Given that 𝑇=2560𝑥𝑦, find the value of 𝑚.

Answer

To solve this problem, we can use the formula for the general term of the binomial expansion to find an alternative expression for 𝑇. We can then equate the two expressions and solve for 𝑚. Recall that the general term of the binomial expansion of (𝑝+𝑞) is given by 𝑇=𝐶𝑝𝑞.

Setting 𝑝=2𝑥, 𝑞=𝑚𝑦, 𝑛=5, and 𝑟=3, we have 𝑇=𝐶(2𝑥)(𝑚𝑦)=10×2𝑚𝑥𝑦=40𝑚𝑥𝑦.

In the question, we are given that 𝑇=2560𝑥𝑦. Hence, we can equate these two expressions for 𝑇 as follows: 2560𝑥𝑦=40𝑚𝑥𝑦.

We can see that both sides of the equation contain the factor 𝑥𝑦, and by equating coefficients, we can write 64=𝑚.

Taking the cube root of both sides of the equation, we obtain 𝑚=4.

In our next example, let us look at how we can use the general term to solve a multistep problem.

Example 5: Using the General Term

If the coefficient of the third term in the expansion of 𝑥14 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 𝑛. Using this, we can solve for 𝑛 and then find the middle term of the expansion. Recall that the general term of the binomial expansion of (𝑝+𝑞) is equal to 𝐶𝑝𝑞𝑟=0,1,,𝑛.for

Since there is a negative sign in the binomial expression, we can begin by writing 𝑥14=𝑥+14.

Note that the expression for the general term begins at 𝑟=0; therefore, to calculate the third term, we need to set 𝑟=2. Substituting 𝑝=𝑥, 𝑞=14, and 𝑟=2, we have 𝐶𝑥14.

Recalling that 𝐶=𝑛𝑛𝑟𝑟, we can rewrite this as 𝑛(𝑛1)2𝑥×116, which simplifies to 𝑛(𝑛1)32𝑥.

Since we are given that the coefficient of this term is 338, we can write 𝑛(𝑛1)32=338.

Multiplying both sides of the equation by 32, we have 𝑛𝑛=132.

If we subtract 132 from both sides of the equation, this leads to the quadratic equation 𝑛𝑛132=0.

We can solve this for 𝑛 by factoring to find (𝑛12)(𝑛+11)=0.

Hence, 𝑛=12 or 𝑛=11. The binomial theorem only applies for the expansion of a binomial raised to a positive integer power. Therefore, 𝑛 must be a positive integer, so we can discard the negative solution and hence 𝑛=12. We can now use this to find the middle term of the expansion. Since 𝑛=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 the seventh term of this expansion. Again, since 𝑟 begins at 𝑟=0, the seventh term in the expansion corresponds to 𝑟=6. Substituting this value into the formula for the general term, we obtain 𝐶𝑥14=9244096𝑥=2311024𝑥.

Therefore, the middle term in the expansion is 2311024𝑥.

If we calculate two consecutive terms in a binomial expansion, we can then find the ratio between them. For the terms 𝑇 and 𝑇, the ratio between them is 𝑇𝑇. We will demonstrate how to calculate this in our next example.

Example 6: Finding the Ratio between Consecutive Terms

Consider the expansion of (8𝑥+2𝑦). Find the ratio between the eighth and the seventh terms.

Answer

Recall that the formula for the general term of the binomial expansion of (𝑝+𝑞) is 𝑇=𝐶𝑝𝑞𝑟=0,1,,𝑛.for

Here, 𝑇 represents the (𝑟+1)th term in the binomial expansion. This means that the seventh term, 𝑇, is obtained using 𝑟=6, and the eighth term, 𝑇, is obtained using 𝑟=7. We can write the general term of the expansion of (8𝑥+2𝑦) by setting 𝑝=8𝑥, 𝑞=2𝑦, and 𝑛=23 as follows: 𝑇=𝐶(8𝑥)(2𝑦)=𝐶×82𝑥𝑦.

As mentioned earlier, we can calculate the seventh term by substituting 𝑟=6: 𝑇=𝐶×8×2𝑥𝑦=𝐶×8×2𝑥𝑦.

Similarly, we can calculate the eighth term by substituting 𝑟=7: 𝑇=𝐶×8×2𝑥𝑦=𝐶×8×2𝑥𝑦.

Therefore, the ratio between the eighth and seventh terms is given by 𝑇𝑇=𝐶×8×2𝑥𝑦𝐶×8×2𝑥𝑦.

Using the rules of exponents, we can simplify this to 𝑇𝑇=𝐶×2𝑦𝐶×8𝑥=𝐶𝑦𝐶×4𝑥=𝐶𝐶×𝑦4𝑥.

Recall that the ratio of consecutive combinations is given by 𝐶𝐶=𝑛𝑟+1𝑟.

Hence, 𝐶𝐶=237+17=177.

Substituting this into the equation above, we have 𝑇𝑇=177×𝑦4𝑥=17𝑦28𝑥.

Therefore, the ratio between the eighth and seventh terms in the binomial expansion is 17𝑦28𝑥.

In the previous 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 𝑇 and 𝑇 of the expansion of (𝑝+𝑞); using the formula for the general term, we can write their ratio as follows: 𝑇𝑇=𝐶𝑝𝑞𝐶𝑝𝑞.

Using the rules of exponents, we can simplify this to 𝑇𝑇=𝐶𝑞𝐶𝑝.

We can now use the formula for ratios of consecutive combinations, 𝐶𝐶=𝑛𝑟+1𝑟, to rewrite this as 𝑇𝑇=(𝑛𝑟+1)𝑟𝑞𝑝.

Formula: The Ratio between Consecutive Terms of a Binomial Expansion

For two consecutive terms 𝑇 and 𝑇 in the expansion (𝑝+𝑞), the ratio between them is 𝑇𝑇=(𝑛𝑟+1)𝑟𝑞𝑝.

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

Example 7: Using Ratios between Consecutive Terms to Solve for Unknowns

Consider the expansion of (𝑎+𝑏), where 𝑎 is positive. Find the values of 𝑎, 𝑏, and 𝑛 given that 𝑇=215040, 𝑇=258048, and 𝑇=215040.

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 𝑇 and 𝑇 in the expansion of (𝑎+𝑏) is given by 𝑇𝑇=(𝑛𝑟+1)𝑟𝑏𝑎.

Substituting 𝑟=6 and 𝑟=5, respectively, along with the values of 𝑇, 𝑇, and 𝑇, we obtain

𝑇𝑇=215040258048=56=(𝑛5)6𝑏𝑎,𝑇𝑇=258048215040=65=(𝑛4)5𝑏𝑎.(1)(2)

By considering the ratio of these two ratios, we can eliminate 𝑎 and 𝑏 and be left with an equation in terms of 𝑛 by dividing equation (1) by equation (2). Hence, ()()=.

This is equivalent to 𝑇𝑇×𝑇𝑇=(𝑛5)6𝑏𝑎×5(𝑛4)𝑎𝑏.

Simplifying this, we get 𝑇𝑇𝑇=5(𝑛5)6(𝑛4).

Substituting in the values of 𝑇, 𝑇, and 𝑇, we have 215040258048=56=5(𝑛5)6(𝑛4).

Dividing both sides by 56 gives 56=𝑛5𝑛4.

Cross multiplying by (𝑛4) and 6, we have 5(𝑛4)=6(𝑛5).

This can be solved as follows: 5𝑛20=6𝑛3020+30=6𝑛5𝑛10=𝑛.

Hence, 𝑛=10.

By substituting our value of 𝑛 into equation (1), we have 56=56𝑏𝑎.

Dividing both sides of the equation by 56 leads to 1=𝑏𝑎. Multiplying both sides of the equation by a gives us 𝑎=𝑏. We can now use the formula for the general term to find the value of 𝑎 and 𝑏 as follows. We can write the general term 𝑇 by substituting 𝑟=4 and set it equal to the given value: 215040=𝐶𝑎𝑏.

Since 𝑎=𝑏, we can rewrite this as 215040=𝐶𝑎.

Recall that 𝐶=𝑛𝑛𝑟𝑟 and therefore 𝐶=1064=10×9×8×74×3×2×1=210.

We have 215040=210𝑎, which simplifies to 1024=𝑎.

Taking the 10th root of both sides of the equation leads to 𝑎=±2. Since we are given that 𝑎 must be positive, we obtain 𝑎=2. We know that 𝑎=𝑏; therefore, 𝑏=2. Our final answer is 𝑎=𝑏=2 and 𝑛=10.

Example 8: Using the Ratio of Consecutive Terms

Consider the binomial expansion of (3+7𝑥) in ascending powers of 𝑥. When 𝑥=6, one of the terms in the expansion is equal to twice its following term. Find the position of these two terms.

Answer

Recall first that the general term of the expansion of (𝑝+𝑞) is 𝑇=𝐶𝑝𝑞𝑟=0,1,,𝑛.for

This leads to an ascending order of powers of 𝑥 when we substitute 𝑎=3 and 𝑏=7𝑥 and systematically increase the value of 𝑟. The question states that when 𝑥=6, one of the terms in the expansion, arranged in ascending powers of 𝑥, is equal to twice its following term. We can write this algebraically as 𝑇=2𝑇.

Hence, 𝑇𝑇=12.

Recall that for the binomial expansion of (𝑝+𝑞), the ratio between consecutive terms is given by 𝑇𝑇=(𝑛𝑟+1)𝑟𝑞𝑝.

Setting 𝑝=3, 𝑞=7𝑥, and 𝑛=28, we can rewrite this as 𝑇𝑇=(28𝑟+1)7𝑥3𝑟.

Since this is equal to a half when 𝑥=6, we can write 12=(29𝑟)7×63𝑟12=(29𝑟)7×2𝑟.

Multiplying both sides of the equation by 2𝑟 gives 𝑟=28(29𝑟)=81228𝑟.

Adding 28𝑟 to both sides of the equation, we have 29𝑟=812.

Dividing through by 29 gives us 𝑟=28. Therefore, the two terms that satisfy the given condition are 𝑇 and 𝑇.

We can also calculate ratios between nonconsecutive terms using similar methods, though the process is a little more involved. We will demonstrate in our final example.

Example 9: Ratios of Nonconsecutive Terms

Consider the expansion of (𝑚𝑥+8), where 𝑚 is a positive constant. Determine the values of 𝑚 and 𝑛, given that the ratio between the coefficients of 𝑇 and 𝑇 is equal to 6374640 and that the ratio between the coefficients of 𝑇 and 𝑇 is equal to 491360.

Answer

One possible approach to this problem would be to directly calculate expressions for the ratios between the coefficients of the twelfth and fourteenth terms and the seventh and ninth terms by applying the formula for consecutive terms. For example, for the ratio between the twelfth and fourteenth terms, we could use the relation 𝑇𝑇=𝑇𝑇×𝑇𝑇.

However, given that we need to calculate two ratios and ultimately form two equations, we will start by deriving an algebraic expression for the ratio of the two terms 𝑇 and 𝑇 and then substitute for the necessary values of 𝑟 (and the other variables) to find the particular expressions for the two ratios.

We can do this by starting with the formula for the ratio of two consecutive terms, 𝑇𝑇, which is 𝑇𝑇=(𝑛𝑟+1)𝑟𝑞𝑝.

We can then form the following equation: 𝑇𝑇×𝑇𝑇=𝑇𝑇.

We can calculate the ratio 𝑇𝑇 by substituting 𝑟+1 into the formula for the ratio of consecutive terms: 𝑇𝑇=(𝑛𝑟+2)𝑟+1𝑞𝑝.

Therefore, 𝑇𝑇=(𝑛𝑟+1)𝑟𝑞𝑝×(𝑛𝑟+2)𝑟+1𝑞𝑝=𝑞(𝑛𝑟+1)(𝑛𝑟+2)𝑝𝑟(𝑟+1).

We have information about the ratio between the coefficients of the twelfth and fourteenth terms and the ratio between the coefficients of the seventh and ninth terms, so we need to start by looking at the reciprocal of our above equation: 𝑇𝑇=𝑝𝑟(𝑟+1)𝑞(𝑛𝑟+1)(𝑛𝑟+2).

In the question, we are told the ratio between the coefficients of the twelfth and fourteenth terms and likewise the ratio between the coefficients of the seventh and ninth terms. By substituting for 𝑝=𝑚𝑥, 𝑞=8, and 𝑟=12, we have 𝑇𝑇=(𝑚𝑥)(12)(13)8(𝑛11)(𝑛10).

Similarly, substituting 𝑟=7 gives 𝑇𝑇=(𝑚𝑥)(7)(8)8(𝑛6)(𝑛5).

As we know the value of the ratio of the coefficients, we can remove the variable from the above ratios and form the following equations:

𝑚(12)(13)8(𝑛11)(𝑛10)=6374640,𝑚(7)(8)8(𝑛6)(𝑛5)=491360.(3)(4)

Dividing equation (3) by equation (4), noting that this is the same as multiplying the equation (3) by the reciprocals of each side of the equation (4), we have noting that this is the same as multiplying the first equation by the reciprocals of each side of the second equation, we have 156𝑚64(𝑛11)(𝑛10)×64(𝑛6)(𝑛5)56𝑚=6374640×136049.

This simplifies to 39(𝑛6)(𝑛5)14(𝑛11)(𝑛10)=22158.

Multiplying by (𝑛11)(𝑛12) and dividing by 3914 yields 11987(𝑛11)(𝑛12)=(𝑛6)(𝑛7).

Multiplying by 87, and multiplying through the parentheses, we have 119𝑛23𝑛+132=87𝑛13𝑛+42.

Gathering all the terms on the left-hand side, we have 32𝑛1606𝑛+12054=0.

We can solve this equation using the quadratic formula that we recall is 𝑛=𝑏±𝑏4𝑎𝑐2𝑎, for the quadratic 𝑎𝑛+𝑏𝑛+𝑐, to find the solutions 𝑛=41 and 𝑛=14716. Since 𝑛 must be a positive integer, we can discard the fractional solution and conclude that 𝑛=41. Finally, we need to substitute 𝑛=41 into either equation (3) or (4) to find 𝑚. We will substitute into equation (4): 𝑚(7)(8)8(416)(415)=491360,561190×64𝑚=491360,11360𝑚=491360.

Hence, 𝑚=49, which gives 𝑚=±7. We are told that 𝑚 is in fact a positive constant that gives us a final answer of 𝑚=7.

Let us finish by recapping a few important concepts from this explainer.

Key Points

  • Using the general term for the binomial expansion often simplifies calculations in which we are only interested in specific terms and their coefficients.
  • The formula for the general term for the binomial expansion (𝑝+𝑞) is 𝑇=𝐶𝑝𝑞𝑟=0,1,,𝑛.for In particular, we should note that the first term corresponds to 𝑟=0. This means that the 𝑘th general term is obtained by using 𝑟=𝑘1 in the general form.
  • Consecutive terms in the binomial expansion (𝑝+𝑞) are related by the formula 𝑇𝑇=(𝑛𝑟+1)𝑟𝑞𝑝.

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