The geometric sequence aₙ = 5·0.75ⁿ⁻¹ decreases. Find the first index n for which aₙ < 0.01.
Sequences & series worksheet with answers
1.Easy
a₁ = 5r = 0.75ε = 0.01
n =
2.Easy
Find the term a40 of the arithmetic sequence −11, −7, −3, 1, …
a₁ = −11d = 4n = 40
a40 =
3.Medium
Exactly one of the four series converges. Click it.
A: Σ 1/√kB: Σ (1 + 1/k)ᵏC: Σ ln k / kD: Σ (−1)ᵏ/√k
- 1A: Σ 1/√k
- 2B: Σ (1 + 1/k)ᵏ
- 3C: Σ ln k / k
- 4D: Σ (−1)ᵏ/√k
4.Medium
Find the sum S = Σk=1∞ 2/(k(k + 4)).
ak = 2/(k(k + 4))
S =
5.Hard
f(x) = cos x is approximated at x = 2 by its Maclaurin polynomial of degree n. With the Lagrange bound M·|x|ⁿ⁺¹/(n + 1)! and M = 1, find the smallest degree n that guarantees an error below 0.001.
f(x) = cos xx = 2M = 1ε = 0.001
n =
6.Hard
Use the Maclaurin polynomial of degree 3 of f(x) = e−x to approximate f(−1.5). Give P₃(−1.5).
f(x) = e−xn = 3x = −1.5
P₃(−1.5) =
Answer key
- 1. n = 23
- 2. a40 = 145
- 3. D: Σ (−1)ᵏ/√k
- 4. S = 1.042
- 5. n = 9
- 6. P₃(−1.5) = 4.188
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