How do you do this question with the given power series?

Step-by-step explanation:

tan⁻¹(x) = ∑ₙ₌₀°° (-1)ⁿ x²ⁿ⁺¹ / (2n+1)

tan⁻¹(1/√3) = ∑ₙ₌₀°° (-1)ⁿ (1/√3)²ⁿ⁺¹ / (2n+1)

tan⁻¹(1/√3) = ∑ₙ₌₀°° (-1)ⁿ (1/√3) (1/√3)²ⁿ / (2n+1)

tan⁻¹(1/√3) = (1/√3) ∑ₙ₌₀°° (-1)ⁿ (1/3)ⁿ / (2n+1)

π/6 = (1/√3) ∑ₙ₌₀°° (-1)ⁿ (1/3)ⁿ / (2n+1)

π = (6/√3) ∑ₙ₌₀°° (-1)ⁿ (1/3)ⁿ / (2n+1)

π = 2√3 ∑ₙ₌₀°° (-1)ⁿ / (3ⁿ (2n+1))

"image"

step-by-step explanation:

in most of these problems, the notation is that the point with a "prime" is the image of the point without. that is, b' is the image of b.

here, there is no dilation or reflection. there is simply translation (with no direction shown). which figure you call the "preimage" and which one you call the "image" is up to you. as a matter of convention, the naming is as described above.