What is the general formula of Binomial Expansion?
The Binomial Theorem is used in expanding an expression raised to any finite power. The binomial theorem states that any non-negative power of binomial (x + y)n can be expanded into a summation of the form , where n is an integer and each n is a positive integer known as a binomial coefficient. Each term in a binomial expansion is assigned a numerical value known as a coefficient. The number of coefficients in the binomial expansion of (x + y)n is (n + 1).
General formula of Binomial Expansion
The general form of binomial expansion of (x + y)n is expressed as a summation function.
where,
n is a positive integer,
x and y are real numbers,
r is an integer such that 0 < r β€ n.
Derivation
The general formula of binomial expansion can be proved using the principle of mathematical induction.
Let,
Step 1: Check the given statement S(n) for n = 1.
So, the result is true for n = 1.
Step 2: Suppose the statement S(n) is true for n = k. So, we get
Step 3: Now, we have to prove that S(k + 1) is true.
(x + y)k+1 = (x + y) (x + y)k
Thus, the result is true for k+1.
Hence, by mathematical induction the result holds true for all positive integers n.
Sample Problems
Problem 1. Find the binomial expansion of (x + 1)4 using the binomial theorem.
Solution:
We have to expand (x + 1)4 using the binomial theorem.
Using the binomial expansion, we get
(x+1)4 = 4C0 x4 + 4C1 x4-1+ 4C2 x4-2 + 4C3 x4-3 + 4C4
=x4 + 4x3 + 6x2 + 4x + 1
Problem 2. Find the binomial expansion of (3x β 2y)4 using the binomial theorem.
Solution:
We have to expand (3x β 2y)4 using the binomial theorem.
Using binomial theorem, we have,
(3x β 2y)4 = 4C0 (3x)4 (2y)0 β 4C1 (3x)3 (2y)1 + 4C2 (3x)2 (2y)2 β 4C3 (3x)1 (2y)3 + 4C4 (3x)0 (2y)4
= 81x4 β 4 (27x3) (2y) + 6 (9x2) (4y2) β 4 (3x) (8y3) + 16y4
= 81x4 β 216x3y + 216x2y2 β 96xy3 + 16y4
Problem 3. Find the binomial expansion of (4x + 9y)5 using the binomial theorem.
Solution:
We have to expand (4x + 9y)5 using the binomial theorem.
Using binomial theorem, we have,
(4x + 9y)5 = 5C0 (4x)5 (9y)0 + 5C1 (4x)4 (9y)1 + 5C2 (4x)3 (9y)2 + 5C3 (4x)2 (9y)3 + 5C4 (4x)1 (9y)4 + 5C5 (4x)0 (9y)5
= 1024x5 + 5 (256x4) (9y) + 10 (64x3) (81y2) + 10 (16x2) (729y3) + 5 (4x) (6561y4) + 59049 y5
= 1024x5 +11520x4y+51840x3y2 +116640x2y3 +131220xy4 +59049y5
Problem 4. Find the binomial expansion of (1 β 6x)6 using the binomial theorem.
Solution:
We have to expand (1 β 6x)6 using the binomial theorem.
Using binomial theorem, we have,
(1 β 6x)6 = 6C0 (3x)0 β 6C1 (6x)1 + 6C2 (6x)2 β 6C3 (6x)3 + 6C4 (6x)4 β 6C5 (6x)5 + 6C6 (6x)6
= 1 β 6 (6x) + 15 (36x2) + 20 (216x3) β 15 (1296x5) + 6 (7776x5) β (46656x6)
= 1 β 36x + 540x2 β 4320x3 + 19440x4 β 46656x5 + 46656x6.
Problem 5. Find the binomial expansion of (1 β 4x + 2x2)3 using the binomial theorem.
Solution:
We have to expand (1 β 4x + 2x2)3 using the binomial theorem.
Using binomial theorem, we have,
(1 β 4x + 2x2)3 = 3C0 (1 β 2x)3 + 3C1 (1 β 2x)2 (3x2) + 3C2 (1 β 2x)(3x2)2 + 3C3 (3x2)3
= (1 β 2x)3 + 9x2 (1 β 2x)2 + 27x4 (1 β 2x) + 27x6
= 1 β 8x3 + 12x2 β 6x + 9x2 (1 + 4x2 β 4x) + 27x4 β 54x5 + 27x6
= 1 β 8x3 + 12x2 β 6x + 9x2 + 36x4 β 36x3 + 27x4 β 54x5 + 27x6
= 1 β 6x + 21x2 β 44x3 + 63x4 β 54x5 + 27x6
Problem 6. Expand (6x + 4)5 using Binomial Theorem.
Solution:
We have to expand (6x + 4)5 using the binomial theorem.
Using binomial theorem, we have,
(6x + 4)5 = 5C0 (6x)5(40) β 5C1 (6x)4 (4) + 5C2 (6x)3 (16) β 5C3 (6x)2 (64)+ 5C4 (6x)1 (256) + 5C5 (6x)0 (1024)
= 7776 x5 + 5(7776x5) + 10 (216x3) (16) + 10 (36x2) (64)+ 5 (6x) (256) + 1024
= 7776x5 + 25920x4 + 34560Γ3 + 23040Γ2 + 7680x + 1024.
Problem 7. Expand (y + 7)5 using Binomial Theorem.
Solution:
We have to expand (y +7)5 using the binomial theorem.
Using binomial theorem, we have,
(y + 7)5 = 5C0 (y)5 (70) β 5C1 (y)4 (7) + 5C2 (y)3 (7)2 β 5C3 (y)2 (73)+ 5C4 (y)1 (74) + 5C5 (y)0 (75)
= y5 + 5(y4)(7) + 10 (y3) (49) + 10 (y2) (343)+ 5 (y) (2401) + 16807
= y5 + 35y4 + 490y3 + 3430y2 + 12005y + 16807
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