Find the term a23 of the arithmetic sequence −7, −5, −3, −1, …
Sequences & series worksheet with answers
1.Easy
a₁ = −7d = 2n = 23
a23 =
2.Easy
The arithmetic sequence 2, 6, 10, 14, … ends with the term 142. How many terms n does it have?
a₁ = 2d = 4aₙ = 142
n =
3.Medium
Apply the ratio test to Σk=1∞ 3ᵏ·k!/kᵏ. Find L = lim |ak+1/ak|.
ak = 3ᵏ·k!/kᵏ
L =
4.Medium
Apply the ratio test to Σk=1∞ 2ᵏ·k!/kᵏ. Find L = lim |ak+1/ak|.
ak = 2ᵏ·k!/kᵏ
L =
5.Hard
f(x) = eˣ is approximated at x = 2 by its Maclaurin polynomial of degree n. With the Lagrange bound M·|x|ⁿ⁺¹/(n + 1)! and M = 8, find the smallest degree n that guarantees an error below 0.0001.
f(x) = eˣx = 2M = 8ε = 0.0001
n =
6.Hard
f(x) = ln(1 + x) is approximated by its Maclaurin polynomial of degree 5 at x = 0.2. Use the Lagrange error bound M·|x|ⁿ⁺¹/(n + 1)! with M = 5! = 120 to find the largest possible error |R₅|.
f(x) = ln(1 + x)n = 5x = 0.2M = 120
|R₅| =
Answer key
- 1. a23 = 37
- 2. n = 36
- 3. L = 1.104
- 4. L = 0.7358
- 5. n = 11
- 6. |R₅| = 1.067 × 10⁻⁵
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