Step-by-step explanation:
a) Use ratio test.
lim(n→∞)│aₙ₊₁ / aₙ│< 1
lim(n→∞)│[x²ᵏ⁺¹ / (2(k+1))!] / [x²ᵏ / (2k)!]│< 1
lim(n→∞)│[x²ᵏ⁺¹ / (2k+2)!] × [(2k)! / x²ᵏ]│< 1
lim(n→∞)│x (2k)! / (2k+2)!│< 1
lim(n→∞)│x / ((2k+2) (2k+1))│< 1
0 < 1
The interval of convergence is (-∞, ∞).
b) f(x) = ∑₎₍₌₀°° x²ᵏ / (2k)!
f'(x) = ∑₎₍₌₁°° 2k x²ᵏ⁻¹ / (2k)!
f'(x) = ∑₎₍₌₁°° x²ᵏ⁻¹ / (2k−1)!
f'(x) = ∑₎₍₌₀°° x²ᵏ⁺¹ / (2k+1)!
c) f(x) + f'(x) = ∑₎₍₌₀°° x²ᵏ / (2k)! + ∑₎₍₌₀°° x²ᵏ⁺¹ / (2k+1)!
Notice f(x) is the sum of even terms (2k) and f'(x) is the sum of odd terms (2k+1). Therefore:
f(x) + f'(x) = ∑₎₍₌₀°° xᵏ / k!
f(x) + f'(x) = eˣ
and get 31.40 cm squared