Mathematics, 16.07.2021 21:50 Crtive6538

Consider the series ∑n=1∞5n2+n. The general formula for the sum of the first n terms is Sn=
. Your answer should be in terms of n.
The sum of a series is defined as the limit of the sequence of partial sums, which means
∑n=1∞5n2+n=limn→∞(
)=
.
Select all true statements (there may be more than one correct answer):
A. Most of the terms in each partial sum cancel out.
B. The series converges.
C. The series is a p-series.
D. The series is a telescoping series (i. e., it is like a collapsible telescope).
E. The series is a geometric series.

Consider the series ∑n=1∞5n2+n.

The general formula for the sum of the first n terms is Sn=
. Yo

Answers: 3

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Answers

Answer from: jettpaquin

Part a. $\displaystyle S_n = 5 - \frac{5}{n + 1}$

Part b. $\displaystyle \lim_{n \to \infty} (5 - \frac{5}{n + 1}) = 5$

Part c. A, B, and D

General Formulas and Concepts:

Algebra I

Terms/CoefficientsFactoring

Pre-Calculus

Partial Fraction Decomposition

Calculus

Limits

Limit Rule [Variable Direct Substitution]: $\displaystyle \lim_{x \to c} x = c$ Limit Property [Addition/Subtraction]: $\displaystyle \lim_{x \to c} [f(x) \pm g(x)] = \lim_{x \to c} f(x) \pm \lim_{x \to c} g(x)$

Sequences

Series

Definition of a convergent or divergent series

Telescoping Series: $\displaystyle \sum^\infty_{n = 1} (b_n - b_{n + 1}) = (b_1 - b_2) + (b_2 - b_3) + (b_3 - b_4) + ... + (b_n - b_{n + 1}) + ...$

Step-by-step explanation:

Step 1: Define

Identify

$\displaystyle \sum^\infty_{n = 1} \frac{5}{n^2 + n}$

Step 2: Rewrite Sum

Factor: $\displaystyle \sum^\infty_{n = 1} \frac{5}{n^2 + n} = \sum^\infty_{n = 1} \frac{5}{n(n + 1)}$ Break up [Partial Fraction Decomposition]: $\displaystyle \frac{5}{n(n + 1)} = \frac{A}{n} + \frac{B}{n + 1}$ Simplify [Common Denominator]: $\displaystyle 5 = A(n + 1) + Bn$ [Decomp] Substitute in n = 0: $\displaystyle 5 = A(0 + 1) + B(0)$ Simplify: $\displaystyle 5 = A$ [Decomp] Substitute in n = -1: $\displaystyle 5 = A(-1 + 1) + B(-1)$ Simplify: $\displaystyle 5 = -B$ Solve: $\displaystyle B = -5$ [Decomp] Substitute in variables: $\displaystyle \frac{5}{n(n + 1)} = \frac{5}{n} + \frac{-5}{n + 1}$ Simplify: $\displaystyle \frac{5}{n(n + 1)} = \frac{5}{n} - \frac{5}{n + 1}$ Substitute in decomp [Sum]: $\displaystyle \sum^\infty_{n = 1} \frac{5}{n^2 + n} = \sum^\infty_{n = 1} \bigg( \frac{5}{n} - \frac{5}{n + 1} \bigg)$

Step 3: Find Sum

Find Sₙ terms: $\displaystyle \sum^\infty_{n = 1} \bigg( \frac{5}{n} - \frac{5}{n + 1} \bigg) = (5 - \frac{5}{2}) + (\frac{5}{2} - \frac{5}{3}) + (\frac{5}{3} - \frac{5}{4}) + (\frac{5}{4} - 1) + ... + ( \frac{5}{n} - \frac{5}{n + 1}) + ...$ Find general Sₙ formula: $\displaystyle S_n = 5 - \frac{5}{n + 1}$ Find Sum [Take limit]: $\displaystyle \sum^\infty_{n = 1} \frac{5}{n^2 + n} = \lim_{n \to \infty} S_n$ Evaluate limit [Limit Rule - Variable Direct Substitution]: $\displaystyle \displaystyle \sum^\infty_{n = 1} \frac{5}{n^2 + n} = 5 + 0$ Simplify: $\displaystyle \sum^\infty_{n = 1} \frac{5}{n^2 + n} = 5$

∴ the sum converges by the Telescoping Series.

Topic: AP Calculus AB/BC (Calculus I/I + II)

Unit: Convergence Tests (BC Only)

Book: College Calculus 10e

Answer from: addisonrausch

(a) Decompose the summand into partial fractions:

$\dfrac5{n^2+n} = \dfrac5{n(n+1)} = \dfrac an+\dfrac b{n+1}$

$\implies 5=a(n+1)+bn=(a+b)n+a$

$\implies a+b=0\text{ and }a=5 \implies b=-5$

$\implies\displaystyle\sum_{n=1}^\infty\frac5{n^2+n} = 5\sum_{n=1}^\infty\left(\frac1n-\frac1{n+1}\right)$

The n-th partial sum for the series is

$S_n = 5\displaystyle\sum_{k=1}^n\left(\frac1k-\frac1{k+1}\right)$

which can be simplified significantly by examinging consective terms in the sum:

$\displaystyle S_n = 5\left(1-\frac12\right) + 5\left(\frac12-\frac13\right) + 5\left(\frac13-\frac14\right) + \cdots + 5\left(\frac1{n-1}-\frac1n\right) + 5\left(\frac1n-\frac1{n+1}\right)$

$\implies S_n = \boxed{5\left(1-\dfrac1{n+1}\right)}$

(b) Using the result of (a), you then get

$\displaystyle\sum_{n=1}^\infty\frac5{n^2+n} = \lim_{n\to\infty}\boxed{5\left(1-\frac1{n+1}\right)} = \boxed{5}$

(c) As shown in (a), the partial sum is simplified because of the reasons given in options A and D, and the result of (b) says that B is also correct.

Answer from: Quest

good job you smart son of a gun you are smarter that einstein. 7

Answer from: Quest

this linear equation would have just one solution.

step-by-step explanation:

the reason why this would have one solution is that it is an expression with just one variable. since that variable is not raised to a power, it will always have just one solution.

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