S=1*2004+2*2003+3*2002+....

Question

S=1*2004+2*2003+3*2002+....+1002*1003=?

Answer 1

我用 SS 代替總和符號 sigma S=SS(k=1 to 1002) k(2005-k)=SS(k=1 to 1002) (-k2+2005k) = -n(n+1)(2n+1)/6+2005n(n+1)/2 given n=1002S=-167(1003)(2005)+2005*501*1003 =334*2005*1003=671,679,010

Answer 2

令R=ab+(a+1)(b-1)+(a+2)(b-2)+......+(a+1001)(b-1001)共1002項
=ab+(ab-a+b-1)+(ab-2a+2b-4)+........+(ab-1001a+1001b-1001^2) [^2是平方]
=1002ab-(1+2+3+.....+1001)a+(1+2+3+......+1001)b-(1+2^2+3^2+.....+1001^2)
=1002ab-[(1+2+3+.....+1001)(a-b)]-[1001(1001+1)(2*1001+1)]/6 {1^2+2^2+3^2+....+n^2=[n(n+1)(2n+1)]/6} 而當a=1.b=2004------
-----1002*1*2004-501*1001*(-2003)-1001*167*2003
=2*2004*501+1001*2003*501-1001*167*2003
=2*(2003+1)*501+1001*2003*501-1001*167*2003
=2*501+2*2003*501+1001*2003*501-1001*167*2003
=671679010----電算機..

Answer 3

S=Σk=1到1002k(2005-k)(為求簡便，以下Σ後省略"k=1到1002")=Σ(2005k-k2)=2005*Σk-Σk2=2005*1002*1003/2-1002*1003*2005/6=671679010

Answer 4

我們先把S簡化(Σ->總合符號)：
Σk(2005-k) [k=1->1002]
=Σ2005k-Σk^2 [k=1->1002]
=2005Σk-Σk^2 [k=1->1002] {Σk=(1+n)n/2、Σk^2=n(n+1)(2n+1)/6，k=1->n}
=2005*(1+1002)1002/2 - 1002(1002+1)(2*1002+1)/6
=671679010