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Expand the following

(n + 2)²

Binomial Expansion

Since (n + 2)² is a binomial
Use the binomial theorem to expand this.

Build binomial expansion

a(x + y)ⁿ = aΣ_{[k = 0 to n]}C(n,k) x^n-ky^k
where

Plug in values

n = 2, x = n, a = 1, and y = 2.
Expanding terms, we get:

k = 0

C(2,0)x^2-0y⁰ = C(2,0)x²y⁰

C(2,0):

C(2,0) = 1
Click to see C(2,0)

Simplify our expanded term:

C(2,0)x²y⁰ = C(2,0)(n)^{2 - 0}(2)⁰

C(2,0)x²y⁰ = (1)(n)²(2)⁰

C(2,0)x²y⁰ = (1)(1)(1n²)(1)

Group constants and powers

C(2,0)x²y⁰ = (1 * 1 * 1 * 1)(n²)

C(2,0)x²y⁰ = 1n²

k = 1

C(2,1)x^2-1y¹ = C(2,1)x¹y¹

C(2,1):

C(2,1) = 2
Click to see C(2,1)

Simplify our expanded term:

C(2,1)x¹y¹ = C(2,1)(n)^{2 - 1}(2)¹

C(2,1)x¹y¹ = (2)(n)¹(2)¹

C(2,1)x¹y¹ = (1)(2)(1n)(2)

Group constants and powers

C(2,1)x¹y¹ = (1 * 2 * 1 * 2)(n¹)

C(2,1)x¹y¹ = 4n

k = 2

C(2,2)x^2-2y² = C(2,2)x⁰y²

C(2,2):

C(2,2) = 1
Click to see C(2,2)

Simplify our expanded term:

C(2,2)x⁰y² = C(2,2)(n)^{2 - 2}(2)²

C(2,2)x⁰y² = (1)(n)⁰(2)²

C(2,2)x⁰y² = (1)(1)(1)(4)

Group constants and powers

C(2,2)x⁰y² = (1 * 1 * 1 * 4)(1)
Anything raised to a 0 power = 1

C(2,2)x⁰y² = 4

Build our answer:

(n + 2)² = 1n² + 4n + 4

************ End Binomial Expansion ********************

Final Answer