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Prove that Σᵣ₌₀ ⁿ 3ʳ ⁿCᵣ = 4ⁿ
Solution:
We will start with the LHS of the given equation that needs to be proved.
LHS = Σᵣ₌₀ ⁿ 3ʳ ⁿCᵣ
= 30 · ⁿC₀ + 31 · ⁿC₁ + 32 · ⁿC₂ + .... + 3n · ⁿCₙ
= ⁿC₀ 1n 30 + ⁿC₁ 1n-1 31 + ⁿC₂ 1n-2 32 + .... + ⁿCₙ 1n-n 3n
Using binomial theorem the above expansion can be written as,
= (1 + 3)n
= 4n
= RHS
Hence proved
NCERT Solutions Class 11 Maths Chapter 8 Exercise 8.1 Question 14
Prove that Σᵣ₌₀ ⁿ 3ʳ ⁿCᵣ = 4ⁿ
Summary:
We have proved that Σᵣ₌₀ ⁿ 3ʳ ⁿCᵣ = 4ⁿ
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