Binomial Theorem

Theory Used

Using the Binomial Theorem: \[ (x+y)^n = \sum_{r=0}^{n} {^nC_r} x^{n-r} y^r \]

Key idea:
Rewrite \(a\) as \(a = (a-b) + b\) so that expansion naturally produces factors of \((a-b)\).

Solution Roadmap

• Rewrite \(a = (a-b)+b\)
• Raise both sides to power \(n\)
• Apply binomial expansion
• Subtract \(b^n\)
• Factor out \((a-b)\)

Factorization Insight

Solution

Rewrite: \[ a = (a-b) + b \]

Raising both sides to power \(n\): \[ a^n = \left[(a-b)+b\right]^n \]

Using binomial expansion: \[ \begin{aligned} a^n &= b^n + {^nC_1}(a-b)b^{n-1} + {^nC_2}(a-b)^2 b^{n-2} + \cdots + (a-b)^n \end{aligned} \]

Subtracting \(b^n\): \[ \begin{aligned} a^n - b^n &= {^nC_1}(a-b)b^{n-1} + {^nC_2}(a-b)^2 b^{n-2} + \cdots + (a-b)^n \end{aligned} \]

Factoring out \((a-b)\): \[ a^n - b^n = (a-b)\left[{^nC_1}b^{n-1} + {^nC_2}(a-b)b^{n-2} + \cdots + (a-b)^{n-1}\right] \]

The bracketed expression is an integer.

\[ \Rightarrow (a-b) \mid (a^n - b^n) \]

Hence proved.

Exam Significance

• Boards: Standard proof question from binomial theorem.
• JEE/NEET: Foundation for:
  • Factor theorem & algebraic identities
  • Divisibility proofs
  • Expression simplification
• Key Insight: Always rewrite cleverly to create a factor.
• Advanced Link: Leads to identity: \(a^n - b^n = (a-b)(a^{n-1} + a^{n-2}b + \cdots + b^{n-1})\)

Theory Used

Using Binomial Expansion: \[ (a\pm b)^n = a^n \pm {^nC_1}a^{n-1}b + {^nC_2}a^{n-2}b^2 \pm \cdots \]

Key identity:
\[ (a+b)^n - (a-b)^n = 2\left[\text{sum of odd power terms}\right] \]

Solution Roadmap

• Let \(a=\sqrt{3}, b=\sqrt{2}\)
• Use symmetry: only odd power terms survive
• Avoid full expansion
• Evaluate required terms efficiently

Symmetry Insight

Solution

Using identity: \[ (a+b)^6 - (a-b)^6 = 2\left[{^6C_1}a^5b + {^6C_3}a^3b^3 + {^6C_5}ab^5\right] \]

Let \(a=\sqrt{3},\ b=\sqrt{2}\), so: \[ a^2=3,\quad b^2=2,\quad ab=\sqrt{6} \]

Compute terms: \[ \begin{aligned} a^5b &= a^4 \cdot ab = 9\sqrt{6} \\ a^3b^3 &= (a^2)(b^2)(ab) = 6\sqrt{6} \\ ab^5 &= b^4 \cdot ab = 4\sqrt{6} \end{aligned} \]

Substitute: \[ \begin{aligned} &= 2\left[6(9\sqrt{6}) + 20(6\sqrt{6}) + 6(4\sqrt{6})\right] \\ &= 2\left[(54 + 120 + 24)\sqrt{6}\right] \\ &= 2(198\sqrt{6}) \\ &= 396\sqrt{6} \end{aligned} \]

Final Answer: \(\;396\sqrt{6}\)

Exam Significance

• Boards: Tests expansion and simplification.
• JEE/NEET: Critical pattern:
  • \((a+b)^n - (a-b)^n\) → only odd terms
  • Reduces 7-term expansion → 3 terms
• Key Insight: Never expand fully → use symmetry.
• Speed Advantage: Saves 70–80% calculation time.

Theory Used

Using identity: \[ (x+y)^n + (x-y)^n = 2\left[\text{sum of even power terms}\right] \]

For \(n=4\): \[ (x+y)^4 + (x-y)^4 = 2(x^4 + 6x^2y^2 + y^4) \]

Solution Roadmap

• Let \(x=a^2,\; y=\sqrt{a^2-1}\)
• Use symmetry identity (avoid full expansion)
• Substitute and simplify

Symmetry Insight

Solution

Let \(x=a^2,\quad y=\sqrt{a^2-1}\)

\[ (x+y)^4 + (x-y)^4 = 2(x^4 + 6x^2y^2 + y^4) \]

Compute: \[ x^4 = a^8,\quad x^2y^2 = a^4(a^2-1),\quad y^4 = (a^2-1)^2 \]

\[ \begin{aligned} &= 2\left[a^8 + 6a^4(a^2-1) + (a^2-1)^2\right] \\ &= 2\left[a^8 + 6a^6 - 6a^4 + a^4 - 2a^2 + 1\right] \\ &= 2\left[a^8 + 6a^6 - 5a^4 - 2a^2 + 1\right] \end{aligned} \]

\[ = 2a^8 + 12a^6 - 10a^4 - 4a^2 + 2 \]

Final Answer: \(2a^8 + 12a^6 - 10a^4 - 4a^2 + 2\)

Exam Significance

• Boards: Tests algebraic simplification.
• JEE: Key identity recognition saves full expansion.
• Insight: Always look for \((x+y)^n + (x-y)^n\).

Theory Used

Using Binomial Expansion: \[ (1-x)^n \approx 1 - nx + \frac{n(n-1)}{2}x^2 \] (first three terms)

Solution Roadmap

• Rewrite number near 1
• Apply binomial approximation
• Use first 3 terms only

Approximation Insight

Solution

\[ (0.99)^5 = (1 - 0.01)^5 \]

Using expansion: \[ \begin{aligned} &\approx 1 - 5(0.01) + \frac{5\cdot4}{2}(0.01)^2 \\ &= 1 - 0.05 + 10(0.0001) \\ &= 1 - 0.05 + 0.001 \\ &= 0.951 \end{aligned} \]

Approximate Value: \(0.951\)

Exam Significance

• Boards: Approximation using binomial expansion.
• JEE: Used in error estimation & numerical methods.
• Key Insight: Works when \(x\) is small.

Theory Used

Use Binomial Theorem: \[ (a-b)^n = a^n - {^nC_1}a^{n-1}b + {^nC_2}a^{n-2}b^2 - \cdots \]

Strategy:
Convert expression into \((a-b)^4\) form to simplify expansion.

Solution Roadmap

• Group terms: \(a = 1 + \frac{x}{2},\; b = \frac{2}{x}\)
• Apply \((a-b)^4\)
• Expand each component separately
• Combine like terms carefully

Structure Insight

Solution

Let \[ a = 1 + \frac{x}{2}, \quad b = \frac{2}{x} \]

Then: \[ (a-b)^4 = a^4 - 4a^3b + 6a^2b^2 - 4ab^3 + b^4 \]

Expanding components: \[ \begin{aligned} a^4 &= 1 + 2x + \frac{3x^2}{2} + \frac{x^3}{2} + \frac{x^4}{16} \\ a^3b &= \frac{2}{x} + 3 + \frac{3x}{2} + \frac{x^2}{4} \\ a^2b^2 &= \frac{4}{x^2} + \frac{4}{x} + 1 \\ ab^3 &= \frac{8}{x^3} + \frac{4}{x^2} \\ b^4 &= \frac{16}{x^4} \end{aligned} \]

Substituting: \[ \begin{aligned} &= a^4 - 4a^3b + 6a^2b^2 - 4ab^3 + b^4 \end{aligned} \]

Final simplification: \[ \begin{aligned} &= \frac{x^4}{16} + \frac{x^3}{2} + \frac{x^2}{2} - 4x + 7 \\ &\quad - \frac{8}{x} + \frac{8}{x^2} - \frac{32}{x^3} + \frac{16}{x^4} \end{aligned} \]

Final Answer: \[ \frac{x^4}{16} + \frac{x^3}{2} + \frac{x^2}{2} - 4x + 7 - \frac{8}{x} + \frac{8}{x^2} - \frac{32}{x^3} + \frac{16}{x^4} \]

Exam Significance

• Boards: Tests multi-step binomial expansion.
• JEE: Important for:
  • Handling mixed powers (positive + negative)
  • Expression simplification
• Key Insight: Smart grouping reduces complexity.
• Common Trap: Missing negative powers of \(x\).

Theory Used

Use identity: \[ (A - B)^3 = A^3 - 3A^2B + 3AB^2 - B^3 \]

Strategy:
Convert 3-term expression into 2-term form for efficient expansion.

Solution Roadmap

• Group terms: \(A = 3x^2 + 3a^2,\; B = 2ax\)
• Apply \((A-B)^3\)
• Expand each term systematically
• Combine like terms carefully

Structure Insight

Solution

Rewrite: \[ (3x^2 - 2ax + 3a^2)^3 = \left[(3x^2 + 3a^2) - 2ax\right]^3 \]

Let: \[ A = 3x^2 + 3a^2,\quad B = 2ax \]

Using identity: \[ (A - B)^3 = A^3 - 3A^2B + 3AB^2 - B^3 \]

Compute each term: \[ \begin{aligned} A^3 &= 27(x^2 + a^2)^3 = 27(x^6 + 3x^4a^2 + 3x^2a^4 + a^6) \\ A^2B &= 18ax(x^4 + 2x^2a^2 + a^4) \\ AB^2 &= 12a^2x^2(x^2 + a^2) \\ B^3 &= 8a^3x^3 \end{aligned} \]

Substitute: \[ \begin{aligned} &= 27(x^6 + 3x^4a^2 + 3x^2a^4 + a^6) \\ &\quad - 3[18ax(x^4 + 2x^2a^2 + a^4)] \\ &\quad + 3[12a^2x^2(x^2 + a^2)] - 8a^3x^3 \end{aligned} \]

Simplifying: \[ \begin{aligned} &= 27x^6 + 81a^2x^4 + 81a^4x^2 + 27a^6 \\ &\quad - 54ax^5 - 108a^3x^3 - 54a^5x \\ &\quad + 36a^2x^4 + 36a^4x^2 - 8a^3x^3 \end{aligned} \]

Combine like terms: \[ \begin{aligned} &= 27x^6 - 54ax^5 + 117a^2x^4 - 116a^3x^3 \\ &\quad + 117a^4x^2 - 54a^5x + 27a^6 \end{aligned} \]

Final Answer: \[ 27x^6 - 54ax^5 + 117a^2x^4 - 116a^3x^3 + 117a^4x^2 - 54a^5x + 27a^6 \]

Exam Significance

• Boards: Multi-step expansion accuracy.
• JEE: Important for:
  • Reducing 3-term expressions → 2-term form
  • Handling symmetric algebraic expressions
• Key Insight: Smart grouping simplifies heavy expansions.
• Common Trap: Missing coefficients during simplification.