How to do it?
What is the least positive integer #n# for which #(n+1)^(1/3)-n^(1/3)<1/12# satisfies?
What is the least positive integer
2 Answers
Explanation:
Note that:
#(x + 1/12)^3 = x^3 + 3(1/12)x^2 + 3(1/12)^2x+(1/12)^3#
#color(white)((x + 1/12)^3) = x^3 + 1/4x^2 + 1/48x+1/1728#
Given:
#(n+1)^(1/3)-n^(1/3) < 1/12#
Add
#(n+1)^(1/3) < n^(1/3)+1/12#
Cube both sides to get:
#color(red)(cancel(color(black)(n)))+1 < color(red)(cancel(color(black)(n)))+1/4 n^(2/3) + 1/48 n^(1/3) + 1/1728#
Subtract
#1728 < 432 n^(2/3) + 36 n^(1/3) + 1#
Subtract
#432 x^2 + 36 x - 1727 > 0#
Rather than solve this properly, let's approximate the positive zero:
#x ~~ sqrt(1727/432) ~~ sqrt(4) = 2#
Then
We find:
#(8+1)^(1/3) - 8^(1/3) = root(3)(9) - 2 ~~ 0.08008 < 0.08bar(3) = 1/12#
8
Explanation:
Let n > 1. Then 1/n < 1.
Use
Here, the absolute values of the terms form a diminishing
sequence. It follows that
And so,
As the value of the expression, for n = 7,
the answer is 8.