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.

Browse more Topics Under Binomial Theorem

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

Solution: 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.

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