Question

How do you do these?

Answer 1

Step-by-step explanation:

(a) The derivative of xⁿ is n xⁿ⁻¹.  Therefore, if ∑ xⁿ = 1 / (1 − x), then:

∑ (n xⁿ⁻¹) = d/dx (1 / (1 − x))

∑ (n xⁿ⁻¹) = 1 / (1 − x)²

(b)(i) ∑ (n xⁿ) = x ∑ (n xⁿ⁻¹) = x / (1 − x)²

(b)(ii) x = 1/9, so the sum is:

1/9 / (1 − 1/9)²

1/9 / (8/9)²

9/64

(c)(i) The derivative of n xⁿ⁻¹ is n (n − 1) xⁿ⁻².  Therefore:

∑ (n (n − 1) xⁿ⁻²) = d/dx (1 / (1 − x)²)

∑ (n (n − 1) xⁿ⁻²) = 2 / (1 − x)³

∑ (n (n − 1) xⁿ) = 2x² / (1 − x)³

(c)(ii) x = 1/4, so the sum is:

2 (1/4)² / (1 − 1/4)³

1/8 / (3/4)³

8/27

(c)(iii) ∑ n² / 2ⁿ

∑ (n² − n) / 2ⁿ + ∑ n / 2ⁿ

2 (1/2)² / (1 − 1/2)³ + 1/2 / (1 − 1/2)²

1/2 / 1/8 + 1/2 / 1/4

4 + 2

6