數學歸納法 - Ans.fyi 參考答案

數學歸納法

[匿名]

2010-09-23 11:49 pm

1(a) 利用數學歸納法,證明對於所有正整數n，1x3 + 2x4 + 3x5 +...+n(n+2) = n(n+1)(2n+7)/ 6

(b) 由此,求下列各式的值.
(i) 1x3+2x4+3x5+...+31x33
(II) 10x12+11x13+12x14+...+1023

回答 (1)

用戶2053

2010-09-24 12:08 am

✔ 最佳答案

a)當 n = 1 , 1(1+2) = 1(1+1)(2(1)+7)/6 = 3 成立 ,設當 n = k 時成立 , 1x3 + 2x4 + 3x5 +...+k(k+2) = k(k+1)(2k+7)/ 6當 n = k+1 ,1x3 + 2x4 + 3x5 +...+k(k+2) + (k+1)(k+3) = k(k+1)(2k+7)/6 + (k+1)(k+3)= (k+1) [k(2k+7)/6 + (k+3)]= (k+1) (2k^2 + 7k + 6k+18) / 6= (k+1) (2k^2 + 13k + 18) / 6= (k+1) (k+2) (2k + 9) / 6= (k+1) (k+2) (2(k+1) + 7) / 6即 n = k + 1 時成立 , 由數學歸納法得證。bi)
1x3+2x4+3x5+...+31x33= 31 * 32 * (2*31 + 7) / 6= 11408bii)
10x12+11x13+12x14+...+1023 = 10x12+11x13+12x14+...+31x33= (1x3+2x4+3x5+...+9x11 + 10x12+11x13+12x14+...+31x33) - (1x3+2x4+3x5+...+9x11)= 11408 - 9*10*(2*9 + 7)/6= 11408 - 375= 11033

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