(x - 1)(x - 2)(x - 3).....(x - 8)(x-9)=a9x9+a8x8+a7x7+.......+a0 則 x7項系數a7=?
回答 (2)
2015-09-26 12:23 pm
a7
= Σ (m,n = -9 to -1 , m<n) mn
= Σ (m,n = 1 to 9 , m<n) mn
= 1*2 + ... + 1*9 + 2*3 + ... + 2*9 + 3*4 + ... + 3*9 + 4*5 + ... + 4*9 + 5*6 + ... + 7*8 + 7*9 + 8*9
= ( (1+2+3+4+5+6+7+8+9)² - (1²+2²+3²+4²+5²+6²+7²+8²+9²) ) / 2
= ( 45² - (9*10*(2*9+1)/6 ) / 2
= 870
= Σ (m,n = -9 to -1 , m<n) mn
= Σ (m,n = 1 to 9 , m<n) mn
= 1*2 + ... + 1*9 + 2*3 + ... + 2*9 + 3*4 + ... + 3*9 + 4*5 + ... + 4*9 + 5*6 + ... + 7*8 + 7*9 + 8*9
= ( (1+2+3+4+5+6+7+8+9)² - (1²+2²+3²+4²+5²+6²+7²+8²+9²) ) / 2
= ( 45² - (9*10*(2*9+1)/6 ) / 2
= 870
2015-09-27 4:40 pm
a7=1*2+1*3+....+1*9+2*3+2*4+...+2*9+....+8*9
而因為(1+2+3+...+9)^2-(1^2+2^2+...+9^2)=2*(1*2+1*3+....+1*9+2*3+2*4+...+2*9+....+8*9)=2a7
=>a7=[(1+2+3+...+9)^2-(1^2+2^2+...+9^2)]/2
=[45^2-(9*10*19/6)]/2
=870
而因為(1+2+3+...+9)^2-(1^2+2^2+...+9^2)=2*(1*2+1*3+....+1*9+2*3+2*4+...+2*9+....+8*9)=2a7
=>a7=[(1+2+3+...+9)^2-(1^2+2^2+...+9^2)]/2
=[45^2-(9*10*19/6)]/2
=870
收錄日期: 2021-04-21 22:31:31
原文連結 [永久失效]:
https://hk.answers.yahoo.com/question/index?qid=20150926024035AAYgX7s