1×2×3+2×3×4+3×4×5...+99×101243+81+27+91+9+3+1+1/3+1/9+1/27+1/81+1/243要用简便方法计算,

来源:学生作业帮助网 编辑:作业帮 时间:2024/05/05 05:26:36

1×2×3+2×3×4+3×4×5...+99×101243+81+27+91+9+3+1+1/3+1/9+1/27+1/81+1/243要用简便方法计算,
1×2×3+2×3×4+3×4×5...+99×101
243+81+27+91+9+3+1+1/3+1/9+1/27+1/81+1/243
要用简便方法计算,

1×2×3+2×3×4+3×4×5...+99×101243+81+27+91+9+3+1+1/3+1/9+1/27+1/81+1/243要用简便方法计算,
n(n+1)(n+2)=[n(n+1)(n+2)(n+3)-n(n+1)(n+2)(n-1)]/[(n+3)-(n-1)]
=[n(n+1)(n+2)(n+3)-n(n+1)(n+2)(n-1)]/4
化简,前后项可以抵消,就剩下首尾项,即[(99*100*101*102)-(0*1*2*3)]/4=25497450
下面一题,题目中多打了一个91吧?在27和9中间,不和规律的
前半段A=243+81+27+9+3+1
A/3=81+27+9+3+1+1/3
A-A/3=243-1/3=242又2/3=2A/3
A=(242又2/3)*3/2=364
后半段B=1/3+1/9+1/27+1/81+1/243
3B=1+1/3+1/9+1/27+1/81
3B-B=1-1/243=242/243=2B
B=121/243
整体A+B=364+121/243
若要加上91,则答案为455+121/243