# 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

= 3^{0 }· ⁿC₀ + 3^{1} · ⁿC₁ + 3^{2} · ⁿC₂ + .... + 3^{n} · ⁿCₙ

= ⁿC₀ 1^{n }3^{0} + ⁿC₁ 1^{n-1 } 3^{1} + ⁿC₂ 1^{n-2 } 3^{2 }+ .... + ⁿCₙ 1^{n-n } 3^{n}

Using binomial theorem the above expansion can be written as,

= (1 + 3)^{n}

= 4^{n}

= 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