Question

binomial theorem write the value of (a+b)^n+(a-b)^n and hence find the value of ( √3+√2)^6-(√3-√2)^6​

Answer 1

Step-by-step explanation:

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Answer 2

The value of (a+b)ⁿ+( a-b)ⁿ =  2[ⁿC₀ aⁿ + ⁿC₂ aⁿ⁻² b² + .....+ ⁿCₙ bⁿ] and ( √3+√2)⁶-(√3-√2)⁶ = 396√6 by using binomial theorem

Given that,

We have to find by using the binomial theorem write the value of (a+b)ⁿ+(a-b)ⁿ and find the value of ( √3+√2)⁶-(√3-√2)⁶.

We know that,

From binomial theorem We have formula

(a+b)ⁿ = ⁿC₀ aⁿ + ⁿC₁ aⁿ⁻¹ b + ⁿC₂ aⁿ⁻² b² + … +ⁿCₙ₋₁ a bⁿ⁻¹ + ⁿCₙ bⁿ  ------>equation(1)

And

(a-b)ⁿ = ⁿC₀ aⁿ - ⁿC₁ aⁿ⁻¹ b + ⁿC₂ aⁿ⁻² b² - … -ⁿCₙ₋₁ a bⁿ⁻¹ + ⁿCₙ bⁿ   -------->equation(2)

Adding equation(1) and equation(2)

(a+b)ⁿ+( a-b)ⁿ =  ⁿC₀ aⁿ + ⁿC₁ aⁿ⁻¹ b + ⁿC₂ aⁿ⁻² b² + … +ⁿCₙ₋₁ a bⁿ⁻¹ + ⁿCₙ bⁿ +  ⁿC₀ aⁿ - ⁿC₁ aⁿ⁻¹ b + ⁿC₂ aⁿ⁻² b² - … -ⁿCₙ₋₁ a bⁿ⁻¹ + ⁿCₙ bⁿ

(a+b)ⁿ+( a-b)ⁿ =  2ⁿC₀ aⁿ + 2ⁿC₂ aⁿ⁻² b² + .....+ 2ⁿCₙ bⁿ

(a+b)ⁿ+( a-b)ⁿ =  2[ⁿC₀ aⁿ + ⁿC₂ aⁿ⁻² b² + .....+ ⁿCₙ bⁿ]

So,

By subtracting equation(1) and equation(2)

(a+b)ⁿ-( a-b)ⁿ =  ⁿC₀ aⁿ + ⁿC₁ aⁿ⁻¹ b + ⁿC₂ aⁿ⁻² b² + … +ⁿCₙ₋₁ a bⁿ⁻¹ + ⁿCₙ bⁿ -  [ⁿC₀ aⁿ - ⁿC₁ aⁿ⁻¹ b + ⁿC₂ aⁿ⁻² b² - … -ⁿCₙ₋₁ a bⁿ⁻¹ + ⁿCₙ bⁿ]

(a+b)ⁿ-( a-b)ⁿ =  2ⁿC₁ aⁿ⁻¹ b + … +2ⁿCₙ₋₁ a bⁿ⁻¹

(a+b)ⁿ-( a-b)ⁿ =  2[ⁿC₁ aⁿ⁻¹ b + … +ⁿCₙ₋₁ a bⁿ⁻¹]

Taking a= √3, b = √2 and n = 6

( √3+√2)⁶-(√3-√2)⁶ = 2[⁶C₁ (√3)⁵ (√2) + ⁶C₃ (√3)³ (√2)³ + ⁶C₅ (√3) (√2)⁵]

= 2[(6)(√3)⁵(√2)+(20)(√3)³(√2)³ + (6)(√3)(√2)⁵]

= 2[54(√6) + 120(√6) + 24 (√6)]

= 396√6

Therefore, (a+b)ⁿ+( a-b)ⁿ =  2[ⁿC₀ aⁿ + ⁿC₂ aⁿ⁻² b² + .....+ ⁿCₙ bⁿ] and ( √3+√2)⁶-(√3-√2)⁶ = 396√6 by using binomial theorem.

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