(1+3+5+···+1999)—(2+4+6+···+1988) = = =

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(1+3+5+···+1999)—(2+4+6+···+1988) = = =
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(1+3+5+···+1999)—(2+4+6+···+1988) = = =
(1+3+5+···+1999)—(2+4+6+···+1988) = = =

(1+3+5+···+1999)—(2+4+6+···+1988) = = =
用等差公式做
【1*1000+(999*1000/2)*2】-【2*994+(994*993/2)*2】
结果自己按计算器吧...

(1+3+5+···+1999)—(2+4+6+···+1988) = = = 1+2-3-4+5+6-7-8+9+10——11-12+······+1997+1998-1999-2000+2001 1 2 3 4 ·····7 ·······10 ········?1 3 5 7 ·····?······?··········290 1 4 9 ·····?······?··········? 找规律:1,1,2,3,5,8,——,21,34,——················ (1994+1992+······+4+2)—(1+3+5+······+1991+1993)=( ) (2+4+6+8+······+100)—(1+3+5+7+······+99)等于几 (1+3+5+······2013)—(2+4+6······+2014)用简便方法哦,这是奥数题. (1+2+3+4+····+1999+2000+1999+···+2+1)÷2000 求(2+5+8······+2000)-(1+4+7······+1999)的值.【简算】 计算(2002+2000+1998+·····+4+2)—(2001+1999+1997+··+3+1) 计算:(—1)+(+2)—3+4—5········+(—2009)+2010—2011+2012= 2[4/3y—(2/3y—1/2)]=3/4y急·································!1 变态数学题·······方程一道············x+2y+3z=1x=y+12x+3y+5z=2方程组············要过程············我解了之后总觉得无解····· 俗语和名句的填空1······,自知之明.2······,行必果.3工欲善其事,······.4临渊羡鱼,······.5非学无以广才,······.6······,福兮祸之所伏.7项庄舞剑,······.8······,人以群 1+2-3-4+5+6-7-8+···+1997+1998-1999 证明1·3·5·····(2n-1)/2·4·6·····2n 1—【1/2】的平方—【1/3】的平方—【1/4】的平方—【1/5】的平方····················会不会有负数?有是第几个? (1+3+5+7+·····+99+101)—(2+4+6+8+····+98+100)