Pure 2.2 - Series 2

Question 1

If the general term, 𝑢ᵣ, of a series can be expressed in the form f(𝑟)-f(𝑟+1), then how can you find Σⁿᵣ₌₁𝑢ᵣ?

Answer 1

If f(𝑟)-f(𝑟+1)
Then Σⁿᵣ₌₁𝑢ᵣ=Σⁿᵣ₌₁(f(𝑟)-f(𝑟+1))
So 𝑢₁=f(1)-f(2), 𝑢₂=f(2)-f(3)… 𝑢ₙ=f(𝑛)-f(𝑛+1).
Thus since all terms except f(1) and -f(𝑛+1) cancel; Σⁿᵣ₌₁𝑢ᵣ=f(1)-f(𝑛+1).

Question 2

Describe the Maclaurin series expansion.

Answer 2

The Maclaurin series expansion of a function f(𝑥) is given by
f(𝑥)=f(0)+f’(0)𝑥+(f’‘(0)/2!)𝑥²+…+(fʳ(0)/𝑟!)𝑥ʳ+…
The series is valid provided that f(0), f’(0), f’‘(0),…, fʳ(0),… all have finite values.