probability question!!thanks!!

(Sum of i² from i=1 to i=n) ≡ n (n+1) (2n+1) / 6, for all natural number n.
[You can prove it by mathematical induction.]

What you think is:
(sum of (i+1)² from i=0 to i=2n)/(n+1)^4
But,
(sum of (i+1)² from i=0 to i=2n)/(n+1)^4
= (sum of i² from i=1 to i=2n+1) / (n+1)^4
= [ (2n+1) (2n+2) (4n+3) / 6 ] / (n+1)^4
= (8n² + 10n + 3) / 3(n+1)³,
which differs from the correct answer!

My calculation is as follows:

Sum of first 2 samples:　0, 1, 2, ... , n-1,　n , n+1, ... , 2n-2, 2n-1, 2n
Freq. of occurrence　 :　1, 2, 3, ... , n　, n+1,　n, ... ,　3　, 2　, 1
(Reasoning:
If the sum = r where 0 ≤ r ≤ n,
permutations are: r+0, (r-1)+1, (r-2)+2, ... , 0+r.
i.e. total no. of terms = r+1.
If the sum = 2n-r where 0 ≤ r ＜ n, i.e. n ＜ n+r ≤ 2n,
permutations are: n+(n-r), (n-1)+(n-r+1), (n-2)+(n-r+2), ... , (n-r)+n.
i.e. total no. of terms = r+1. )

Similarly,
Sum of last 2 samples:　0, 1, 2, ... , n-1,　n , n+1, ... , 2n-2, 2n-1, 2n
Freq. of occurrence　 :　1, 2, 3, ... , n　, n+1,　n, ... ,　3　, 2　, 1

Therefore,
Sum of first 2 samples & sum of last
　2 samples are BOTH:　0 , 1 , 2 , ... , n-1,　n　 , n+1, ... , 2n-2, 2n-1, 2n
Freq. of occurrence　 :　1², 2², 3², ... , n²　, (n+1)²,　n² , ... ,　3² ,　2² , 1²

Sum of "frequency of occurrence"
= [1² + 2² + 3² + ... + n² + (n+1)²] + [1² + 2² + 3² + ... + n²]
= [(n+1) (n+2) (2n+3) / 6] + [n (n+1) (2n+1) / 6]
= (n+1)/6 . [(n+2) (2n+3) + n (2n+1)]
= (n+1) (2n² + 4n + 3) /3
Therefore,
probability = [(n+1) (2n² + 4n + 3) /6 ] / (n+1)^4
　　　　　= (2n² + 4n + 3) / [3 (n+1)³]
　which is same as the correct answer as suggested.

Your idea is correct, but your formula is wrong.

You sum from i = 0 to 2n. However, think about when can the first two sum and last two sum equal 2n?? only 1 possible choice, not 2n+1 choices!!
so actually it is:
i=0: no. of ways: 1^2
i=1: no. of ways: 2^2
i=2: no. of ways: 3^2
...
i=n: no. of ways: (n+1)^2
**i=n+1: no. of ways: n^2
**i=n+2: no. of ways: (n-1)^2
...
**i=n: no. of ways: 1^2

Therefore it should be: 1+2^2+...+n^2 + (n+1)^2 + n^2 + (n-1)^2+...+1^2

using formula Sum i^2 (i=1 to n) = n(n+1)(2n+1)/6, the above becomes"

S(n) + S(n+1)
=n(n+1)(2n+1)/6 + (n+1)(n+2)(2n+3)/6
= 2(n+1)(2n^2+4n + 3) /6

divide by (n+1)^4 and simplify will give you the correct answer.

2007-01-27 12:02:46 補充：
the last one should be...**i=2n: no. of ways: 1^2