Introduction
In early mathematics, students learn how to square and cube binomials like (a+b)(a+b) and (a−b)(a-b). This knowledge helps in calculations such as:
982=(100−2)2,9993=(1000−1)398^2 = (100 - 2)^2, \quad 999^3 = (1000 - 1)^3
However, when dealing with higher powers such as (95)5(95)^5 or (101)6(101)^6, repeated multiplication becomes tedious.
To simplify these calculations, the Binomial Theorem was introduced. Traditionally, it involves permutations and combinations (nCr^nC_r), which are typically introduced in higher classes (Class 11 and above).
But what if there was a simpler approach that anyone—from a fifth grader to a PhD student—could use without any knowledge of complex formulas?
Here’s a step-by-step method that makes binomial expansion easy and intuitive.
The Formula
For any binomial (a+b)n(a+b)^n, the expansion follows this pattern:
(a+b)n=anb0+nan−1b11+n(n−1)an−2b22+⋯+a0bn(a+b)^n = a^n b^0 + \frac{n a^{n-1} b^1}{1} + \frac{n(n-1) a^{n-2} b^2}{2} + \dots + a^0 b^n
This structure allows easy calculation of binomial expansions without the need for combinations.
Step-by-Step Process:
Let's expand (a+b)2(a+b)^2 as an example:
Start with the highest power of aa and the lowest power of bb:
(a+b)2=a2b0(a+b)^2 = a^2 b^0
Since anything raised to the power of zero is 1, we simplify:
a2a^2
For the second term:
Multiply the previous power of aa by the exponent nn (which is 2).
Reduce the power of aa by 1.
Increase the power of bb by 1.
Divide by the number of preceding terms.
2a1b11=2ab\frac{2a^1 b^1}{1} = 2ab
Continue this process until the power of aa becomes 0:
(a+b)2=a2+2ab+2a0b22(a+b)^2 = a^2 + 2ab + \frac{2a^0 b^2}{2} =a2+2ab+b2= a^2 + 2ab + b^2
More Examples:
Example: (a+b)3(a+b)^3
(a+b)3=a3b0+3a2b11+3×2a1b22+3×2×1a0b33(a+b)^3 = a^3 b^0 + \frac{3a^2 b^1}{1} + \frac{3 \times 2 a^1 b^2}{2} + \frac{3 \times 2 \times 1 a^0 b^3}{3}=a3+3a2b+3ab2+b3= a^3 + 3a^2 b + 3ab^2 + b^3
Example: (a+b)4(a+b)^4
(a+b)4=a4b0+4a3b11+4×3a2b22+4×3×2a1b33+4a0b44(a+b)^4 = a^4 b^0 + \frac{4a^3 b^1}{1} + \frac{4 \times 3 a^2 b^2}{2} + \frac{4 \times 3 \times 2 a^1 b^3}{3} + \frac{4a^0 b^4}{4}=a4+4a3b+6a2b2+4ab3+b4= a^4 + 4a^3 b + 6a^2 b^2 + 4ab^3 + b^4
Expanding Binomials with a Negative Term
For (a−b)n(a-b)^n, replace bb with −b-b in the formula:
(a−b)n=an+(−b)0+nan−1(−b)11+n(n−1)an−2(−b)22+⋯+a0(−b)n(a-b)^n = a^n + (-b)^0 + \frac{n a^{n-1} (-b)^1}{1} + \frac{n(n-1) a^{n-2} (-b)^2}{2} + \dots + a^0 (-b)^n
This follows the same pattern, except alternate terms will change signs.
Conclusion
Mathematics is all about recognizing patterns. With this simple approach, anyone can expand binomials without relying on complex formulas involving permutations and combinations.
So next time you see an expression like (102)5(102)^5 or (99)6(99)^6, you won’t need a calculator—you’ll just follow the pattern and simplify it step by step!
Hence, we say, MATHS IS ALL ABOUT PATTERN!
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