初中數學競賽裂項法

裂項法基本類型計算：.一一?-[答疑編號5721030101]99[答疑編號5721030101]99【答案】ioo【解答】99W01+1)72-1-1?11—總結公式：””：?U■.■?>.?一常見變形:T1.T1.（改變間隔）計算：1乂32x411

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97x9958x100[答疑編號5721030102]14651【答案】19800；十97^9999x100；十97^9999x1001111111111111,=—*[(1--)+(--1-(232435469799)+(98~i00)]111122991001465119EOO2x3x4+2x3x4+++57x98x9998x99x10041x0.（增加乘數）計算：一[答疑編號5721030103]

4949【答案】1泥00【解答】原式三4-22原式三4-22小499-97100-98、+十)97x98x9998x95x100-x[()+(-)-I1-(-)]21x22x32x33x498x9999x1001一H(——)21^299乂1。049491S8C0.(整數裂項)計算：1X2+2X3+3X4+……+99X100[答疑編號5721030104]【答案】333300【解答】原式=-x[C1k2x3-0x]k2)+(2x3x^-1x2x3)+--+(99x100x101-98x99x100)]=1x95x100x101=3333003進一步思考:.計算：1X2X3+2X3X4+3X4X5+……+98X99X100SE[答疑編號5721030105]【答案】24497550【解答】=-x[(lx2x3x4-0xlx2>>3)+(2x3x4?<5-^2x3?<4)+-+(98x499x100x101-97x98x99x100)]=-x98x99x]OOxl014=2450x9999=2450'<(10000-1)=24497550再進一■步5.計算:F+2n+3*H卜丹[答疑編號5721030106]—xy?x(用+1/【答案】4原式=:0句聲2十0十。乂2制3+2）十--十［（再一1）次界河5十1）十對=1x2x3h-2x3x4-i---十（打一1）x算工（用十1）十1十2十--十用—x-1)xwx(w+1)y(m+2)+工乂用乂(置+1)42=—M理父(盟+1)x[5—l)X^+2)+2]41-1=-xhjx(?+5j4靈活的例題產22產223210W之十-+一+…二‘一1」9」二[答疑編號5721030107]■產嚅原式二k2十上十￡4U33x55x72000原式二k2十上十￡4U33x55x720002H-—十1999x2001-x(K—+1+—41x33x5—+--F1-F5x71999x20011111111111——x[l000+—x(1-—4H4233557二迎期+\網)422001小風200119992001)17.乂―7.乂―1乂2M45乂6]十一十j1999*2QQQ百=」100010011002[答疑編號百=」100010011002[答疑編號5721030108]1999【答案】2000【解答】1114kF…T1x23x45x61999x20005【答案】2000【解答】1114kF…T1x23x45x61999x200056)[19992000J11+…+1199g200U,-2.1+1+..+【242000JW00_1111=iooiW021003"2000那么田_且二’一'='

WOO2C0020008,]2x200713x200611720081x2009+8,]2x200713x200611720081x2009++2x20083x20072003x11-I2009x1比較20091月與2010x3的大?。鄞鹨删幪枺鄞鹨删幪?721030109][答案]2010kB>2Q09x月「、….20092009200920092009y.A=kF+--I1x20082x20073x20062008x11+2008,2+2007.3+2006,2008十1=++■+lx2C082x20073x20062003x1TOC\o"1-5"\h\z,111111r=[++—+十一H-HF+12C0S22007320062008,Zd111、=2x(1+―+―+■■-+)232008同理，21HoM3=2攵(14…+^-)2320092可見2。10吊6=2009x上十二所以2010xB>200**.2009

23419—■++?口?+9.求3!41512川的近似值