The formula of Binomial theorem has a great role to play as it helps us in finding binomial’s power. The procedure is made much easier as it doesn’t have to pass through the boring multiplying process.
Moreover, the usage of the formula also helps in determining the middle and general term in the binomial expansion of the mathematical expression.
In this section, you’ll be seeing the middle and the general term when the binomial (a +b)ⁿ is represented in the expanded form with the use of the binomial theorem.
The Formula of Binomial Theorem In The Form of General Term
By the use of the formula of Binomial theorem, it’s known that there are (n + 1) terms present in the binomial expansion belonging to (a+b)ⁿ.
Now, let’s state that T1, T2, T3, T4, T₅ … Tn+1 are the 1st, 2nd, 3rd, 4th, 5th, ….. (n + 1)th terms, in a respective manner, regarding the (a+b)ⁿ expansion.
T1 = (n 0)aⁿ
T2 = (n1)aⁿ-1.b
T3 = (n2)aⁿ–2.b2
T4 = (n3)aⁿ–3.b3…..
Tn+1 = (n / n)bn
On generalization, we have general term’s formula:
Tr+1 = (n / r)aⁿ–r. br
where 0 ≤ r ≤ n. Let’s look at an example now.
Question: Find the 4th term in the binomial expansion of (4x–2y)⁶.
a = 4x, b = – 2y, and n = 6. From the formula above, we’re having
Tr+1 = (n / r)a^n–r. b^r
To determine the 4th term, T4, r = 3. Thus,
T4 = T3+1 = (7 / 3) (4x)^6–3.(−2y)^3
= 6.5.4 / 1.2.3. (64x³). (-8y³)
= – 10,240.x³.y³
Hence, the 4th term in the expansion of (4x–2y)^6 = – 10,240x³y³
The formula for Binomial Theorem In The Form of Middle Term
If you want to expand the expression (a+b)ⁿ & ‘n’ is an even no. and therefore (n + 1) refers to an odd number. This means that the binomial expansion will consist of terms related to odd numbers.
In this condition, the middle term of binomial theorem formula will be equal to (n / 2 + 1)th term.
Let’s say if you expand (x+y)², therefore, the middle term results in the form the (2 / 2 + 1) which is equal to 2nd term. (n2 + 1)th term is also represented as the (n+2 / 2)th term.
When (a+b)ⁿ is the expression, you try to expand & ‘n’ refers to an odd number, and accordingly, (n + 1) will be equal to an even number. Thus, there exist 2 middle terms namely, (n+1 / 2)th term & (n+3 / 2)th term.
Let’s say, if you’re expanding the expression, (x+y)³, later the middle terms are equal to (3+1 / 2) = 2nd term & (3+3 / 2) = 3rd term.
Example : Find the middle terms in the expansion of (x / 4 + 2y)^11.
Given: n = 11, which is an odd number & we are having 2 middle terms which are: (11+1 / 2) = 6th term & (11+3 / 2) = 7th term.
We are also having,
a = x / 4, b = 2y, and n = 11.
As it’s known that,
Tr+1 = (n / r)a^n–r. b^r
To find the 6th term,
T6 , r = 5. Thus,
T6 = T5+1 = (11 / 5)(x / 4)^11–5. (2y)5
= 22.214.171.124.7 / 126.96.36.199.5. (x⁶ / 4096) . (32y⁵)
= 231 / 64. x⁶. y⁵
T7 = T6+1 = (11 / 6)(x / 4)^11–6. (2y)⁶
= 188.8.131.52.7.6 / 184.108.40.206.5.6 .(x⁵ / 1024) . (729y⁶)
= 231 / 8 . x⁵. y⁶
Thus, (x / 4 + 2y)^11 expansion’s middle terms = 231 / 8 .x⁵. y⁶