數列A為項數是104的等差數列且級數和是2015;數列B為項數是104的等比數列且級數和是520;兩數列的順序乘積和(如同向量內積)為20000;求兩數列的逆序乘積和?

2015-09-13 7:22 am

回答 (1)

2015-09-13 1:39 pm
✔ 最佳答案
令:
數列A的首項為 a , 公差為 d ;
數列B的首項為 b , 公比為 r ;
兩數列的逆序乘積和為 S

分別將數列A, B 按數列順序寫成104維向量,
並令 B' 為 B向量的逆順序向量, 則:
A = ( a , a+d , a+2d , a+3d , ..... , a+102d , a+103d )
B = ( b , br , br^2 , br^3 , ..... , br^102 , br^103 )
B' = ( br^103 , br^102 , br^101 , ..... , br , b )

Σ A = 2015 , 所以:
104( 2a + 103d ) / 2 = 2015
2a + 103d = 2015*2/104 = 38.75
d = ( 38.75 - 2a ) / 103

Σ B = 520 , 所以:
b( 1 - r^104 ) / (1-r) = 520
1 + r + r^2 + ..... + r^102 + r^103 = 520/b

20000
= A.B
= ab + (a+d)br + (a+2d)br^2 + (a+3d)br^3 + ..... + (a+102d)br^102 + (a+103d)br^103
= ab(1+r+r^2+.....+r^102+r^103) + db(r+2r^2+3r^3+.....+102r^102+103r^103)
= ab(520/b) + db(r+2r^2+3r^3+.....+102r^102+103r^103)
= 520a + db(r+2r^2+3r^3+.....+102r^102+103r^103)

S
= A.B'
= abr^103 + (a+d)br^102 + (a+2d)br^101 + (a+3d)br^100 + ..... + (a+102d)br + (a+103d)b
= ab(r^103+r^102+.....+r+1) + db(r^102+2r^101+3r^100 +.....+102r+103)
= ab(520/b) + db(r^102+2r^101+3r^100 +.....+102r+103)
= 520a + db(r^102+2r^101+3r^100 +.....+102r+103)

令:
p = r + 2r^2 + 3r^3 + ..... + 102r^102 + 103r^103
q = r^102 + 2r^101 + 3r^100 + ..... + 102r + 103
則:
20000 = 520a + dbp
S = 520a + dbq

p + q
= 103 + 103r + ..... + 103r^102 + 103r^103
= 103( 1 + r + r^2 + ..... + r^102 + r^103 )
= 103 * 520/b
= 53560/b

S + 20000
= 1040a + db(p+q)
= 1040a + db*53560/b
= 1040a + 53560d
= 1040a + 53560( 38.75 - 2a )/103
= 1040a + 520( 38.75 - 2a )
= 20150

S = 20150 - 20000 = 150

Ans: 150


收錄日期: 2021-05-02 14:06:57
原文連結 [永久失效]:
https://hk.answers.yahoo.com/question/index?qid=20150912232237AAkCIze

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