1x2十2x3十3x4推算过程

孙刷殷3197求解,Sn=1x2+2x3+3x4+…+n(n+1)=?跪求,要过程 -
喻牲穆13147196912 ______ 裂项相消法. 原式:1x2+2x3+3x4+...+n(n+1) =1^2+1+2^2+2+3^2+3+...+n^2+n =1+2+...+n+(1^2+2^2+...+n^2) =(1+n)n/2+n(n+1)(2n+1)/6 =n(n+1)/2*(1+(2n+1)/3) =n(n+1)(2n+5)/6 欢迎采纳!我帮你!!

孙刷殷31971x2十2x3十3x4十4x5十5x6 , -
喻牲穆13147196912 ______[答案] 1x2十2x3十3x4十4x5十5x6 = 2 + 6+12+20+30 = 70 无量寿佛,佛说苦海无涯回头是岸! 施主,我看你骨骼清奇,器宇轩昂,且有慧根,乃是万中无一的武林奇才. 潜心修习,将来必成大器,吾手中正好有一本宝典,欲赠于施主 鄙人有个小小的考验...

孙刷殷31971x2+2x3+3x4+......+19x20 -
喻牲穆13147196912 ______ 【解答】原式*3可以得到如下变形 1*2*3+2*3*3+3*4*3+……+19*20*3 =1*2*3+2*3*(4-1)+3*4*(5-2)+……+19*20*(21-18) =1*2*3-1*2*3+2*3*4-2*3*4+3*4*5+……-18*19*20+19*20*21 =19*20*21 由此可以知道原式=19*20*21÷3=2660

孙刷殷31971x2+2x3+3x4+4x5+5x6+6x7+7x8+8x9= -
喻牲穆13147196912 ______ 1x2 + 2x3 + 3x4 + ...... + n(n+1) = 1/3 * n(n+1)(n+2) 1x2+2x3+3x4+4x5+5x6+6x7+7x8+8x9 = 1/3 * 8 * 9 * 10 = 240

孙刷殷31971x2+2x3+3x4+...+10x11=? -
喻牲穆13147196912 ______ 展开全部1x2+2x3+3x4+...+10x11=2+6+12+20+30+42+56+72+90+110=440

孙刷殷31971x2+2x3+3x4......+99x100得数的简算方法 -
喻牲穆13147196912 ______ 1x2+2x3+3x4......+99x100 =2(1x2/2+2x3/2+3x4/2......+99x100/2) =2[C(2,2)+C(3,2)+C(4,2)+.....+C(100,2)] =2[C(3,3)+C(3,2)+C(4,2)+.....+C(100,2)] 连续利用公式C(n,m)+C(n,m-1)=C(n+1,m) =2*C(100,3) =2*100*99*98/6 =323400

孙刷殷31973(1x2十2x3十3x4…十2011x2012)怎么计算? -
喻牲穆13147196912 ______[答案] n(n+1)=n²+n 原式=3(1²+1+2²+2+……+2011²+2011) =3(1²+2²+3²+……2011²+1+2+3+.2011) 1+2+3+n=n(n+1)/2 1²+2²+3²+……n²=n*(n+1)*(2n+1)/6 带入,原式的值可求

孙刷殷31971x2+2x3+3x4+···+10x11 -
喻牲穆13147196912 ______ 1*2=1*3(1*2*3-0*1*2)2*3=1/3(2*3*4-1*2*3)...............10*11=1/3(10*11*12-9*10*11) 所以1x2+2x3+3x4+···+10x11=1/3(10*11*12-0*1*2)=440

孙刷殷31971x2+2x3+3x4+......+100x101=? -
喻牲穆13147196912 ______ x2+2x3+3x4+......+100x101= 的通项公式An=n*(n+1)=n²+n 前n项和Sn=(1²+2²+……+n²)+(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

孙刷殷31971x2十2x3十3x4十4x5……+19x20=? -
喻牲穆13147196912 ______ ∵公式:1x2十2x3十3x4十4x5……+n*(n+1)=n*(n+1)(n+2)/3∴1x2十2x3十3x4十4x5……+19x20=19*20*21/3=2660