1. The sequence a(n) satisfies a(1) = 3 and

a(n) = [a(n-1)-1]/a(n-1)+1

for all n ≥ 2. Find a(100).

(I'm sorry about the way I formatted this question.

2. One of the roots of 2x^3 − 9x^2 + 13x + k = 0 is x = 2. Find the other two roots.

Chadly Jun 17, 2024

#1**0 **

Problem 1:

To find a(100), we first need to find a(2), a(3), a(4), and so on until we reach a(100) by applying the recursive formula given.

a(1) = 3

To find a(2):

a(2) = [a(1)-1]/a(1)+1

a(2) = [3-1]/3+1

a(2) = 2/4

a(2) = 1/2

To find a(3):

a(3) = [a(2)-1]/a(2)+1

a(3) = [1/2-1]/1/2+1

a(3) = -1/2 / 3/2

a(3) = -1/3

Continuing on in this fashion, we find:

a(4) = 1/2

a(5) = 1

a(6) = 0

It's clear that the sequence oscillates between a positive number and zero.

Therefore, for n ≥ 5, a(n) = 1 for odd n and a(n) = 0 for even n.

Since 100 is even, a(100) = 0.

Rangcr897 Jun 17, 2024

#2**-1 **

Problem 2:

To find the other two roots, we can use polynomial division or synthetic division to divide the given polynomial by (x - 2) since x = 2 is a root.

Using synthetic division:

2|2 -9 13 k

2 -4 18

___________

2 -7 9 k

Therefore, the polynomial can be factored as:

2x^3 − 9x^2 + 13x + k = (x - 2)(2x^2 - 7x + 9)

Now, to find the other two roots, we can solve the quadratic equation 2x^2 - 7x + 9 = 0 using the quadratic formula:

x = (-(-7) ± √((-7)^2 - 4*2*9)) / 2*2

x = (7 ± √(49 - 72)) / 4

x = (7 ± √(-23)) / 4

Since the discriminant is negative, the quadratic equation has two complex roots:

x = (7 ± √23i) / 4

Therefore, the other two roots are:

x = (7 + √23i) / 4

x = (7 - √23i) / 4

Rangcr897 Jun 17, 2024