expansion of (1-x-2x^2)^n = 1 - nx + [(1/2n(n-1)-2n]x^2 + [2n(n-1)-1/6n(n-1)(n-2)]x^3 + ...
taking x = 1/8 ,evaluate square root of 6 to 4 significant figure.
i'm having problem with the square root of 6,can't find its relationship with 1/8...
can anybody help me......
Problem in binomial expansion?
2010-12-04 8:11 am
回答 (1)
2010-12-04 9:12 am
✔ 最佳答案
Set x = 1/8, then 1-x-2x^2 = 1 - 1/8 - 1 / 32 = 32 - 4 - 1 / 32 = 27 /32 = 6 * 9/64 = 6*(3/8)^2.
Set n = 1/2 and you'll get (1-x-2x^2)^n = 3/8 * sqrt(6). Got it?
收錄日期: 2021-05-03 14:25:00
原文連結 [永久失效]:
https://hk.answers.yahoo.com/question/index?qid=20101204001158AAj7CzL