2. Write a recursive formula for the sequence 5, 18, 31, 44, 57, ... Then find the next term.

Mathematics

2 answers:

ludmilkaskok [199]2 years ago

8 0

Answer: a6=70

Step-by-step explanation:

we know that

In the Arithmetic Sequence the formula is:

an= a1 + d(n−1)

where

a1 is the first term

d is the common difference

n is the number of terms

in this problem

a1=5

a2=18

a3=31

a4=44

a5=57

d=a2-a1 or a3-a2 or a4-a3 or a5-a4

d=57-44----> 13

find a6

a6= a1 + d(n−1)

n=6

d=13

a1=5

a6=5+13*(6-1)----> a6=5+13*5----> a6=70

Wittaler [7]2 years ago

3 0

Answer:

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I need help with part b. I feel like there’s a catch, I want to do the first derivative test, however, I feel like there is a be

Sladkaya [172]

Answer:

The fifth degree Taylor polynomial of g(x) is increasing around x=-1

Step-by-step explanation:

Yes, you can do the derivative of the fifth degree Taylor polynomial, but notice that its derivative evaluated at x =-1 will give zero for all its terms except for the one of first order, so the calculation becomes simple:

$P_5(x)=g(-1)+g'(-1)\,(x+1)+g"(-1)\, \frac{(x+1)^2}{2!} +g^{(3)}(-1)\, \frac{(x+1)^3}{3!} + g^{(4)}(-1)\, \frac{(x+1)^4}{4!} +g^{(5)}(-1)\, \frac{(x+1)^5}{5!}$

and when you do its derivative:

1) the constant term renders zero,

2) the following term (term of order 1, the linear term) renders: $g'(-1)\,(1)$ since the derivative of (x+1) is one,

3) all other terms will keep at least one factor (x+1) in their derivative, and this evaluated at x = -1 will render zero

Therefore, the only term that would give you something different from zero once evaluated at x = -1 is the derivative of that linear term. and that only non-zero term is: $g'(-1)= 7$ as per the information given. Therefore, the function has derivative larger than zero, then it is increasing in the vicinity of x = -1