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 1^n-1  3¹ + ⁿC₂ 1^n-2  3² + .... + ⁿCₙ 1^n-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