+22 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.
+856 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.
+856 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
