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Binomial Theorem Numerical Applications

CBSE Class 11 MathsBinomial Theorem🟒 Free Lesson

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Binomial Theorem Numerical Applications

[MathDefinition title="Numerical Applications"] The binomial theorem can be used for approximating values, finding remainders, and solving numerical problems without full expansion. [/MathDefinition]

[MathKeyFormula title="Key Approximations"]

  1. (1 + x)^n β‰ˆ 1 + nx for small x (|x| << 1)
  2. (1 + x)^n β‰ˆ 1 + nx + n(n-1)x^2/2 for better approximation
  3. (1 - x)^n β‰ˆ 1 - nx for small x
  4. (1 + x)^n * (1 + y)^n β‰ˆ 1 + n(x + y) for small x, y [/MathKeyFormula]

[MathNote title="When to Use Approximation"] Use binomial approximation when x is very small compared to 1. The error decreases as more terms are included. [/MathNote]

[MathExample title="Example 1: Approximating a Root"] Problem: Find the approximate value of (1.02)^10.

Solution: (1.02)^10 = (1 + 0.02)^10

Using binomial approximation (first two terms): β‰ˆ 1 + 10(0.02) = 1 + 0.2 = 1.2

Using first three terms: β‰ˆ 1 + 10(0.02) + 10(9)(0.02)^2/2 = 1 + 0.2 + 45(0.0004) = 1 + 0.2 + 0.018 = 1.218

The exact value is approximately 1.21899, so the approximation is very close. [/MathExample]

[MathExample title="Example 2: Approximating a Fraction"] Problem: Find the approximate value of (0.99)^5.

Solution: (0.99)^5 = (1 - 0.01)^5

Using binomial approximation: β‰ˆ 1 - 5(0.01) + 10(0.01)^2 - 10(0.01)^3 = 1 - 0.05 + 0.001 - 0.00001 = 0.95099

The exact value is approximately 0.95099, so the approximation is excellent. [/MathExample]

[MathExample title="Example 3: Finding Remainder"] Problem: Find the remainder when 2^100 is divided by 7.

Solution: 2^100 = (2^3)^33 * 2 = 8^33 * 2

8 ≑ 1 (mod 7), so 8^33 ≑ 1^33 = 1 (mod 7)

Therefore, 2^100 ≑ 1 * 2 = 2 (mod 7)

The remainder when 2^100 is divided by 7 is 2. [/MathExample]

[MathExample title="Example 4: Estimating Value"] Problem: Estimate the value of 1.01^20 using binomial theorem.

Solution: 1.01^20 = (1 + 0.01)^20

Using binomial approximation: β‰ˆ 1 + 20(0.01) + 190(0.01)^2 + 1140(0.01)^3 = 1 + 0.2 + 0.019 + 0.00114 = 1.22014

The exact value is approximately 1.22019, so the estimate is very accurate. [/MathExample]

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Binomial Theorem Numerical Applications

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