cosh(2r) – 16. (a) Show that sinh?(z) =2(b) Find the nth term of the following sequence{1,8, 10, 26, 46, · - - }for all n > 1.

The two problems are as follows.

Answered: cosh(2r) – 1 6. (a) Show that sinh (z)…

Calculus: Early Transcendentals

8th Edition

ISBN:9781285741550

Author:James Stewart

Publisher:James Stewart

Chapter1: Functions And Models

Section: Chapter Questions

Problem 1RCC: (a) What is a function? What are its domain and range? (b) What is the graph of a function? (c) How...

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I have done a bunch of these of problems, but this one step looks wrong in the solution for this problem. https://www.bartleby.com/solution-answer/chapter-62-problem-71e-calculus-volume-2-17th-edition/9781938168062/in-the-following-exercises-express-each-series-as-a-rational-function-71-n11-x3-2n1/b209f360-2097-11e9-8385-02ee952b546e On the very first step when I distribute the (x-3)^(2n-1) exponent in the denominator to be (x-3)^2n and (x-3)^-1, the (x-3)^-1, should flip up to the numerator (not stay in the denominator as (x-3)), but it doesn't according to the answer key. And the answer is right but that first step LOOKS wrong. Can you help explain that exponent distribution?
Th 2n +2 Diverges to - Converges to 1. Converges to 8. Diverges to +a0
Lemma 4. Let {yn} be a positive solution of equation (1.1) with the corresponding sequence {z„} e So for n > N. Then: (i) (1– p)zn N; (ii) {zn¢n} is increasing for all n > N. Proof. Assume that {yn} is a positive solution of equation (1.1) with the corres- ponding sequence {zn} E So. Then z, is positive, Zn 2 yn, and Yn = Zn – Pnyo(n) 2 (1– p)zn, n>N> no, so (i) is proved. It easy to see that z, E So implies lim b,(Az,)“ = 0; n00 otherwise we would eventually have Az, > 0 contradicting z, E So. Similarly, lim a„A(b,(Azn)“) = 0. A summation of equation (1.1) then yields 9szs+1 Zn+1 LIs. s=n s=n s=n Summing once more, we obtain as t=s s=n or Azn 2-Zn+1Qn. Hence, A(zn&n) = PnAzn + Zn+1Ao, > Zn+1(A0n – OnQn) =0 since {0,} is a solution of the difference equation (Ao, – Qnºn) = 0. Therefore, {z,9n} is increasing and this completes the proof of the lemma.
Mark each sequence {n} having the property that lim infan> inf{n: n € N}. n+∞0 U U 0 U Xn = = {cos(nπ/3)} (-1)"n n+1 Xn = In = 1 Xn= n 1 when n is odd, and n = 1+ - when n is even. n xn = (-1)" n n- (-1)"

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cosh(2r) – 1 6. (a) Show that sinh (z) = (b) Find the nth term of the following sequence {1,8, 10, 26, 46, · … - } for all n 2 1.