HELP!!!!!!binomial同M.I.

用戶24800

2010-08-18 11:17 pm

1. prove that n^(1/n) < 3^(1/3) for any natural number n greater or equal to 4
by M.I

2..(ax+1/x^2)^n is expanded in descending powers of x, where n is a positive
interger and a>0. If the fourth term of the expansion is independent of x and is
equal to 21/2, find the values of n and x.

3.Find the constants a ,b and c such that for any natural number m,
m^3= a．mC3 + b ．mC2 + c ．mC1 and hence find 1^3+2^3+3^3+.........+n^3=?

回答 (1)

六呎將軍

2010-08-19 2:30 am

✔ 最佳答案

1)

圖片參考：http://i388.photobucket.com/albums/oo325/loyitak1990/Aug10/CrazyMI1.jpg

2) The fourth term is:

nC3 (ax)n-3 (1/x2)3 = 21/2

nC3 an-3 xn-9 = 21/2

Hence n = 9 and then:

9C3 a6 = 21/2

a6 = 1/8

a = 1/√2

3) a x mC3 + b x mC2 + c x mC1

= am(m - 1)(m - 2)/6 + bm(m - 1)/2 + cm

= a(m3 - 3m2 + 2m)/6 + b(m2 - m)/2 + cm

= am3/6 + (-a/2 + b/2)m2 + (a/3 - b/2 + c)m

Thus

a/6 = 1, a = 6

-a/2 + b/2 = 0, b = 6

a/3 - b/2 + c = 0, c = 1

Finally:

13 = 1C1
23 = 6 x 2C2 + 2C1
33 = 6 x 3C3 + 6 x 3C2 + 3C1
43 = 6 x 4C3 + 6 x 4C2 + 4C1
.
.
n3 = 6 x nC3 + 6 x nC2 + nC1

Summing up and simplifying, we have:

13 + 23 + ... + n3 = n2(n + 1)2/4

2010-08-19 08:29:28 補充：
Q1)
http://i388.photobucket.com/albums/oo325/loyitak1990/Aug10/CrazyMI1.jpg

Q2) independent of x 的意思為 x 的 0 次方, 即無論 x 為多少都是一樣.
n 則為 9

參考: Myself

👍 0 👎 0