Question

Given cot θ = 2√2, the sum of the infinite series 1 + 2 (1 – sin θ) + 3 (1 – sin θ)² + 4 (1 – sin θ)³ + ..... is

(a) 6√2 

(b) 8 

(c) 9 

(d) 8√2

Answer 1

(c) 9

Let S_∞ = 1 + 2 (1 – sin θ) + 3 (1 – sin θ)² + 4(1 – sin θ)³ + ..... ∞ 

⇒ S_∞ = 1 + 2a + 3a² + 4a³ + ..... ∞              (where a = 1 – sin θ) 

a S_∞ = a + 2a² + 3a³ + ..... ∞ 

⇒ S_∞ – a S_∞ = 1 + a + a² + a³ + ..... ∞ 

(1 – a) S_∞ = \(\frac{1}{1-a}\)

⇒ S_∞ = \(\frac{1}{(1-a)^2}\) = \(\frac{1}{(1-1+ \text{sin}\,\theta)^2}\)

= \(\frac{1}{\text{sin}^2\,\theta}\) = cosec² θ

⇒ S_∞ = cosec²θ = 1 + cot²θ = 1 + (2√2)² = 1 + 8 = 9.