Beginning & Intermediate Algebra
A sequence is defined by the formula an=n2−1a_n = n^2 - 1 for 1≤n≤51 \(\le\) n \(\le\) 5. List all the terms of this sequence.
Write an explicit formula for the arithmetic sequence whose first term is a1=−12\(\displaystyle\) a_1 = -\(\frac{1}{2}\) and whose common difference is d=14\(\displaystyle\) d = \(\frac{1}{4}\).
Consider the sequence defined by an=(−1)n⋅n22n\(\displaystyle\) a_n = (-1)^n \(\cdot\) \(\frac{n^2}{2^n}\) for n≥1n \(\ge\) 1. Which classification best describes the behavior and properties of this sequence?
Simplify the explicit formula an=3+(−2)⋅(n−1)a_{n}=3+\(\left\)(-2)\(\cdot\)(n-1\(\right\)) into standard linear form.
Two arithmetic sequences are given by an=6+4(n−1)a_{n}=6+4(n-1) and bn=2+6(n−1)b_{n}=2+6(n-1). Find the smallest positive integer nn such that an=bna_n = b_n.
Find the first nn for which the arithmetic sequence with a1=50a_1 = 50 and d=−7d = -7 becomes nonpositive (an≤0)\(\left\)(a_{n}\(\le\)0\(\right\)).
Which explanation correctly justifies why 0!=10! = 1 using the recursive identity n!=n⋅(n−1)!n! = n\(\cdot\)(n−1)!?
Simplify (n+2)!n!\(\frac{(n + 2)!}{n!}\) for a positive integer nn. Express your answer in the simplest multiplicative form.
Compute 12!10!\(\frac{12!}{10!}\) by choosing the most efficient strategy.
The sequence 2,4,8,16,...2, 4, 8, 16, ... is best classified as:
Compute the 1010th term of the geometric sequence with a1=81a_1 = 81 and r=−13r=-\(\frac\)13.
A student claims the sequence given by sn=5⋅2ns_n = 5·2^n is geometric with common ratio r=2r = 2 and first term a1=5a_1 = 5. Evaluate the correctness of that claim with respect to the standard an=a1⋅rn−1a_{n}=a_1\(\cdot\) r^{n-1} indexing, and if needed, provide the correct a1a_1 for the student's formula or rewrite sns_n to match the standard form.
