数学题

kangson

<!--QuoteBegin-kavin+12.01.2005, 21:23 --></div><table border='0' align='center' width='95%' cellpadding='3' cellspacing='1'><tr><td><b>QUOTE</b> (kavin @ 12.01.2005, 21:23 )</td></tr><tr><td id='QUOTE'><!--QuoteEBegin--> 都错，不过我也曾算出来11 <!--QuoteEnd--> </td></tr></table><div class='postcolor'> <!--QuoteEEnd--><br> you are right, i have had a mistake

(n1, n2, ..., nk)(transponiert)  mal  (n1, n2, ..., nk)

zansan

(n1+n2+n3+...+n99)^2<br />=n1*n1+n1*n2+...+n2*n1+n2*n2+.....n98*n99<br />=[n1*n2+n1*n3+n1*n4+...+n2*n3+n2*n4+...+......+n98*n99]+[n2*n1+n3*n1+n3*n2+n4*n1+n4*n2+.....+n99*n1+n99*n2+...+n99*n97]+(n1^2+n2^2+...+n99^2)<br />=2*(n1*n2+n1*n3+n1*n4+...+n2*n3+n2*n4+...+......+n98*n99)+(n1^2+...+n99^2)<br />=2*A+99<br />其中(n1^2+...+n99^2)=99,      (n1+n2+...+n99)=2*k+1<br />A=n1*n2+n1*n3+n1*n4+...+n2*n3+n2*n4+...+......+n98*n99<br /><br />(n1+n2+...+n99)^2=1,9,25,49,...,(2*k+1)^2<br /><br />当(n1+n2+...+n99)&gt;10时,A&gt;0.<br />A最小正值为(11^2-99)/2=11<br />A的最小值=(1-99)/2=49

zansan

不好意思,A最小值=-49