What is the third term in the expansion of (x + 2)⁶?

Solution:

Given (x + 2)⁶

We use binomial theorem to get the expansion

(a + b)ⁿ = ∑(k = 0 to n) ⁿC_k (a_{n - k} × b^k)

Here, a = x and b = 2, n = 6

(x + 2)⁶ = ∑(k=0 to n) ⁶C_k (x)^{6 - k} × (2)^k

(x + 2)⁶ = 6! /6!0! (x)⁶(2)⁰ + 6!/(6 - 1)!1! (x)⁵(2)¹ + 6!/(6 - 2)!2! (x)⁴(2)² + 6!/(6 - 3)!3! (x)³(2)³ + 6!/(6 - 4)!4! (x)²(2)⁴ + 6!/(6 - 5)!5! (x)¹(2)⁵

(x + 2)⁶ = 1 × (x)⁶(2)⁰ + 6 × (x)⁵(2)¹ + 15 × x⁴(2)² + 20x³(2)³ + 15 × x²(2)⁴ + 6x¹(2)⁵ + 1 × x⁰(2)⁶

(x + 2)⁶ = x⁶ + 12x⁵ + 60x⁴ + 160x³ + 240x² + 192x + 64

Coefficient of the third term is 60

What is the third term in the expansion of (x + 2)⁶?

Summary:

The third term in the expansion of (x + 2)⁶ is 60.