求救！高一數學統計問題

1.1. Xbar=?N1 = 20, Xbar1 = 75, S1 = 8N2 = 30, Xbar2 = 80, S2 = 6
ΣX1 = N1*Xbar1= 20*75= 150
ΣX2 = N2*Xbar2= 30*80= 240
ΣX = ΣX1 + ΣX2= 150 + 240= 390
Xbar = ΣX/N= 390/50= 78
1.2. S = ?S1^2 = ΣX1^2/N1 - Xbar1^2S2^2 = ΣX2^2/N2 - Xbar2^2=>ΣX1^2 = N1*(S1^2 + Xbar1^2)ΣX2^2 = N2*(S2^2 + Xbar2^2)=>S^2 = ΣX^2/N - Xbar^2= (ΣX1^2 + ΣX2^2)/(N1 + N2) - Xbar^2= [N1*(S1^2 + Xbar1^2) + N2*(S2^2 + Xbar2^2)]/(N1 + N2) - Xbar^2= [20*(64 + 5625) + 30*(36 + 6400)]/50 - 78^2= (20*5689 + 30*6436)/50 - 6084= 53.2
S = 7.293833

2.幫我出一題最基本的數學歸納法證明，然後幫我付解答過程
數學歸納法證明: Sn = 1+2+3+...+n = n(n+1)/2
n = 1 => 左邊 = 1右邊 = 1*2/2 = 1 = 左邊
n = 2 => 左邊 = 1 + 2 = 3右邊 = 2*3/2 = 3 = 左邊
.......設 n = k - 1, S<k-1> = (k-1)k/2 成立則 n = k => Sk = (k-1)k/2 + k = k[(k-1)/2 + 1]= k(k-1+2)/2= k(k+1)/2所以 Sn = n(n+1)/2 成立

2015-03-30 08:36:58 補充：
3.數學歸納法證明: Sn = 1^6 + 2^6 + 3^6 +...+ n^6

= n(n+1)(2n+1)(3n^4+6n^3-3n+1)/42


n = 1 => 左邊 = 1

右邊 = 1*2*3*(3+6-3+1)/42

= 6*7/42

= 1

= 左邊

= 成立



n = 2

左邊 = 1 + 64 = 65

右邊 = 2*3*5*(48+48-6+1)/42

= 30*91/42

= 30*13/6

= 5*13

= 65

= 左邊 成立

......

2015-03-30 08:38:34 補充：
設 n = k - 1

S = (k-1)k(2k-1)[3(k-1)^4 + 6(k-1)^3 - 3(k-1) + 1]/42

= k(k-1)(2k-1)(3k^4-6k^3+3k+1)/42

成立


則 n = k

Sk = k(k-1)(2k-1)(3k^4-6k^3+3k+1)/42 + k^6

= k*{(k-1)(2k-1)(3k^4-6k^3+3k+1) + 42k^5}/42

= k(6k^5+21k^5+21k^4-7k^2+1)/42

= k(k+1)(6k^5+15k^4+6k^3-6k^2-k+1)

2015-03-30 08:39:10 補充：
= k(k+1)(2k+1)(3k^4+6k^3-3k+1)/42

也成立


所以 Sn = n(n+1)(2n+1)(3n^4+6n^3-3n+1)/42 成立

合併標準差 = √{ [20(75² + 8²) + 30(80² + 6²) - 50(78)²]/50 }

2015-03-29 01:49:13 補充：
數學歸納法證明 －＞ 在 Google 一搜尋就有很多例題了。　

2015-03-29 17:03:41 補充：
哈哈，我剛剛比麻辣老師慢了，讓我把我的解答放在意見欄吧，請看以下的連結：

https://s.yimg.com/rk/HA00430218/o/627762676.png