Hello. Please answer the attached Discrete Mathematics question correctly. Do not copy the answers already given on Chegg or elsewhere. I just want your own personal answer. Show all your work. *If you answer correctly and do not copy from outside sources, I will provide a Thumbs Up. Thank you. A Silicon Valley billionaire purchases 3 new cars for his collection at the end of every month. Let an denote the number of cars he hasafter n months. Let a = 23.(a) What is ag?(b) If he pays $50 each month to have each car maintained, what is the total amount that he has paid for maintenance after 2 years?No need to calculate the actual number. Instead give a closed form (without the summation) mathematical expression for thenumber. Note that he purchases the new cars at the end of each month, so during the first month, he is only maintaining 23 cars.

It is given that Person B purchases 3 cars every month. We denote the total number of cars after nmonths by an . Initial number of cars,i.e. a0=23 It is given that Person B spends $50 per month to have a car maintained. Arithmetic Progression:- an=a1+(n-1)d where a1 is the first term,an is nth term and d is the common difference between terms. Sum of arithmetic progression:- Sn=n2[2a+(n-1)d] where Sn is the sum of n terms, a is the first term of the arithmetic progression and d is the common difference between terms. a) We want to find a8. We have, a1=a0+3=23+3=26 a2=a1+3=26+3=29 Hence, we get an arithmetic progression an with d=3 and first term a1=26given by, an=a1+(n-1)3 ⇒an=26+3n-3 ⇒an=23+3n Now putting n=8,we get, a8=23+(3×8) ⇒a8=23+24 ⇒a8=47 b) We want to find the total amount which he has paid for maintenance after 2 years. i.e. 24 months It costs $50×23=$1150 for the maintenance of all cars in the first month. It costs $50×26=$1300 for the maintenance of all cars in the second month. Hence, we get an arithmetic progression with b1=1150 and d1=150. The total amount Person B has paid for maintenance is given by the sum of this arithmetic progression. The sum of this arithmetic progression is given by, Sn=n2[2b1+(n-1)d1] ⇒Sn=n2[2(1150)+(n-1)150] ⇒Sn=n2[2300+150n-150]⇒Sn=n2[2150+150n]which is the required expression. Substituting n=24 in the above equation,we get S24 which is the total amount Person B has paid for maintenance after 24 months.

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A Silicon Valley billionaire purchases 3 new cars for his collection at the end of every month. Let a denote the number of cars he has after n months. Let ao = 23. (a) What is ag? (b) If he pays $50 each month to have each car maintained, what is the total amount that he has paid for maintenance after 2 years? No need to calculate the actual number. Instead give a closed form (without the summation) mathematical expression for the number. Note that he purchases the new cars at the end of each month, so during the first month, he is only maintaining 23 cars.

A Silicon Valley billionaire purchases 3 new cars for his collection at the end of every month. Let a denote the number of cars he has after n months. Let ao = 23. (a) What is ag? (b) If he pays $50 each month to have each car maintained, what is the total amount that he has paid for maintenance after 2 years? No need to calculate the actual number. Instead give a closed form (without the summation) mathematical expression for the number. Note that he purchases the new cars at the end of each month, so during the first month, he is only maintaining 23 cars.

Advanced Engineering Mathematics

10th Edition

ISBN:9780470458365

Author:Erwin Kreyszig

Publisher:Erwin Kreyszig

Chapter2: Second-order Linear Odes

Section: Chapter Questions

Problem 1RQ

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