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1x2+2x3+...+100x101=

x2+2x3+3x4++100x101= 的通项公式an=n*(n+1)=n+n 前n项和sn=(1+2+……+n)+(1+2+……+n) =n(n+1)(2n+1)/6+n(n+1)/2 所以 1x2+2x3+3x4++100x101=s(100) =100*101*201/6+100*101/2=343400

1x2+2x3+3x4++100x101=[(1x2+2x3+3x4++100x101)x3]/3=[1x2x(3-0)+2x3x(4-1)+3x4x(5-2)+.+100x101x(102-109)]/3=(-0x1x2+1x2x3-1x2x3+2x3x4-2x3x4+3x4x5-.-109x100x101+100x101x102)/3=100x101x102/3=3434001x2+2x3+3x4++n(n+

(1)1x2+2x3+3x4++100x101=1/3*(100*101*102-99*100*101+99*100*101-0*1*2)=1/3*102*100*101=343400(2)1x2+2x3+3x4+n(n+1)==1/3*[n*(n+1)*(n+2)-(n-1)*n*(n+1)-0*1*2]=1/3n(n+1)(n+2)(3)1x2x3+2x3x4+3x4x5++n(n+1)(n+2)=1/4[n(n+1)(n+2)(n+3)-(n-1)n(n+1)(n+2)+(n-1)n(n+1)(n+2)..-0*1*2*3)]=1/4n(n+1)(n+2)(n+3)

1*2=1*1+1 1*2+2*3=1*1+1+2*2+2 1x2+2x3+3x4=1*1+1+2*2+2+3*3+3 以此类推 1x2+2x3+3x4++100x101 =1x1+1+2x2+2+3x3+3++100x100+100 =[1x1+2x2+3x3+4x4++100x100]+[1+2+3+4+..+100] 又因为 1^2+2^2+3^2++n^2=[n*(n+1)*(2n+1)]/6 1+2+3+4+..+n=[(1+n)n]/2 所以 原式=[100*(100+1)*(100*2+1)]/6+(1+100)*100/2 答案自己算吧我没带计算器

化简为(2-1)*2+(3-1)*3+(101-1)*100 =2+3+4+100-2-3-4-100 接下来请你自己动脑子啦,很简单吧!!!

1*2+2*3++100*101=1/3*(1*2*3-0*1*2) +1/3*(2*3*4-1*2*3) ++1/3*(100*101*102-99*100*101) =1/3*1*2*3-1/3*0*1*2 +1/3*2*3*4-1/3*1*2*3 ++1/3*100*101*102-1/3*99*100*101= -1/3*0*1*2 +1/3*100*101*102=0+343400=343400愿

1x2+2x3+3x4+.+100x101= 的通项公式An=n*(n+1)=n+n 前n项和Sn=(1+2+……+n)+(1+2+……+n) =n(n+1)(2n+1)/6+n(n+1)/2 所以 1x2+2x3+3x4+.+100x101=S(100) =100*101*201/6+100*101/2=34340

利用两个公式1+2+3+.+n=n(n+1)/21+2+3+.+n=n(n+1)(2n+1)/61x2+2x3+3x4++100x101 =1x1+2x2+3x3++100x100+(1+2+3+.+100)=100x101x201/6+(1+100)*100/2=343400

原式=(1*2+1)+(2^2+2)+(3^2+3)+……+(99^2+99)+(100*100+100)=1/6*100*(100+1)*(100*2+1)+(100+1)*100/2=100*101*201÷6+5050=剩下的自己算

我来回答343400

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