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Binomial Theorem

General and Middle Term

The Binomial theorem formula helps us to find the power of a binomial without having to go through the tedious process of multiplying it. Further use of the formula helps us determine the general and middle term in the expansion of the algebraic expression too. In this article, we will look at some simple ways to find the middle term and general term when a binomial \( (a + b)^n \) is expanded using the binomial theorem.

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Binomial Theorem Formula – General Term

Binomial theorem formula

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By the Binomial theorem formula, we know that there are (n + 1) terms in the expansion of \( (a + b)^n \). Now, let’s say that \( T_1 \), \( T_2 \), \( T_3 \), \( T_4 \), … \( T_{n + 1} \) are the first, second, third, fourth, … (n + 1)th terms, respectively in the expansion of \( (a + b)^n \). Therefore,

\( T_1 \) = \( {n} \choose {0} \)\( a^n \)
\( T_2 \) = \( {n} \choose {1} \)\( a^{n – 1} \).b
\( T_3 \) = \( {n} \choose {2} \)\( a^{n – 2} \).\( b^2 \)
\( T_4 \) = \( {n} \choose {3} \)\( a^{n – 3} \).\( b^3 \)
…..
\( T_{n + 1} \) = \( {n} \choose {n} \)\( b^n \)

Generalizing it, we have the formula for the general term:

\( T_{r + 1} \) = \( {n} \choose {r} \)\( a^{n – r} \)\( b^r \)

where 0 ≤ r ≤ n. Let’s look at an example now.

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Example 1

Find the fourth term in the expansion of \( (3x – y)^7 \).

Solution. In this example, a = 3x, b = – y, and n = 7. From the above formula, we have
\( T_{r + 1} \) = \( {n} \choose {r} \)\( a^{n – r} \)\( b^r \)

To find the fourth term, \( T_4 \), r = 3. Therefore,
\( T_4 \) = \( T_{3 + 1} \) = \( {7} \choose {3} \)\( (3x)^{7 – 3} \).\( (- y)^3 \)
= \( \frac {7 . 6 . 5}{1 . 2 . 3} \)\( (81x^4) \).\( (- y)^3 \) = – 2835\( x^4 \)\( y^3 \)

Hence, the fourth term in the expansion of \( (3x – y)^7 \) = – 2835\( x^4 \)\( y^3 \)

Binomial Theorem Formula – Middle Term

When you are trying to expand \( (a + b)^n \) and ‘n’ is an even number, then (n + 1) will be an odd number. Which means that the expansion will have odd number of terms. In this case, the middle term will be the (\( \frac {n}{2} \) + 1)th term.

For example, if you are expanding \( (x + y)^2 \), then the middle term will be the (\( \frac {2}{2} \) + 1) = 2nd term. (\( \frac {n}{2} \) + 1)th term is also denoted as (\( \frac {n + 2}{2} \))th term.

When you are trying to expand \( (a + b)^n \) and ‘n’ is an odd number, then (n + 1) will be an even number. Hence, there are two middle terms: (\( \frac{n + 1}{2} \))th term and (\( \frac {n + 3}{2} \))th term.

For example, if you are expanding \( (x + y)^3 \), then the middle terms will be, (\( \frac {3 + 1}{2} \)) = 2nd term and (\( \frac {3 + 3}{2} \)) = 3rd term. Let’s look at an example now.

Example 2

Find the middle term/s in the expansion of (\( \frac {x}{2} \) + 3y)9.

Solution. In this example, since n (= 9) is odd, we have two middle terms namely,
(\( \frac {9 + 1}{2} \)) = 5th term and (\( \frac {9 + 3}{2} \)) = 6th term.

We also have,
a = \( \frac {x}{2} \), b = 3y, and n = 9.

We know that,
\( T_{r + 1} \) = \( {n} \choose {r} \)\( a^{n – r} \)\( b^r \)

To find the fifth term, \( T_5 \), r = 4. Therefore,
\( T_5 \) = \( T_{4 + 1} \) = \( {9} \choose {4} \)\( (\frac {x}{2})^{9 – 4} \)\( (3y)^4 \)
= \( \frac {9 . 8 . 7 . 6}{1 . 2 . 3 . 4} \)(\( \frac {x^5}{32} \))(\( 81y^4 \)) = \( \frac {5103}{16} \)\( x^5 \)\( y^4 \).

Similarly,
\( T_6 \) = \( T_{5 + 1} \) = \( {9} \choose {5} \)\( (\frac {x}{2})^{9 – 5} \)\( (3y)^5 \)
= \( \frac {9 . 8 . 7 . 6 . 5}{1 . 2 . 3 . 4 . 5} \)(\( \frac {x^4}{16} \))(\( 243y^5 \)) = \( \frac {15309}{8} \)\( x^4 \)\( y^5 \).

Therefore, the middle terms in the expansion of (\( \frac {x}{2} \) + 3y)are \( \frac {5103}{16} \)\( x^5 \)\( y^4 \) and \( \frac {15309}{8} \)\( x^4 \)\( y^5 \).

More Solved Examples for You

Question 1: Find the coefficient of \( x^6 \) in the expansion of \( (x + 2)^9 \).

Answer : We know that the binomial expansion of \( (a + b)^n \) is,
\( T_{r + 1} \) = \( {n} \choose {r} \)\( a^{n – r} \)\( b^r \)

Now, the (r + 1)th term in the expansion of \( (x + 2)^9 \) will be:
\( T_{r + 1} \) = \( {9} \choose {r} \)\( (x)^{9 – r} \)\( (2)^r \)

From the above equation, we can deduce that the coefficient of the x-term is \( {9} \choose {r} \)\( (2)^r \)

Next, we need to find the coefficient of \( x^6 \). Hence,
\( x^6 \) = \( (x)^{9 – r} \)
Or, 6 = 9 – r, therefore r = 3.

Using this value of ‘r’,
The coefficient of \( x^6 \) = \( {9} \choose {3} \)\( (2)^3 \)
= \( \frac {9 . 8 . 7}{1 . 2 . 3} \).(8) = 672.

Solved Questions for You

Question 1: What is the use of Binomial theorem?

Answer: The Binomial theorem formula assists us in finding out the power of a binomial without having to go through the dreary method of multiplying it. Further, use of the formula will assist us in determining the general and middle term in the expansion of the algebraic expression too.

Question 2: What is a binomial expression?

Answer: A binomial refers to a mathematical expression that contains two terms, which we must separate through addition or subtraction. To add binomials, you need to combine like terms to get your answer. To multiply binomials, you need to make use of the distributive property. More often than not, you will not get a binomial answer with multiplication

Question 3: What is a binomial factor?

Answer: Binomial factors refer to polynomial factors that have exactly two terms. Binomial factors are easy to solve, and the roots of the binomial factors are the same as the roots of the polynomial. Thus, the first step to find the roots is to factor a polynomial.

Question 4: What is the square of a binomial?

Answer: The square of a binomial refers to the sum of the square of the first terms, twice the product of the two terms, and finally the square of the last term.

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