Question

计算：24×(1/2×3+1/4×5+……+1/(20×21))

Answer 1

1/[n(n+1)]=1/n -1/(n+1)
1/(1²+2²+...+n²)=1/[n(n+1)(2n+1)/6]
=6/[n(n+1)(2n+1)]
=6[1/n + 1/(n+1) -4/(2n+1)]
24×[1/(2×3)+1/(4×5)+...+1/(20×21)] -[1/1²+1/(1²+2²)+...+1/(1²+2²+...+10²)]
=24×(1/2-1/3+1/4-1/5+...+1/20-1/21)-6(1/1+1/2-4/3+1/2+1/3-4/5+...+1/10+1/11-1/21)
=24×(1/2+1/4+...+1/20)-24×(1/3+1/5+...+1/21)-6(1/1+1/2+1/2+1/3+...+1/10+1/11)+24(1/3+1/5+...+1/21)
=24×(1/2+1/4+...+1/20) -6[1+2×(1/2+1/3+...+1/10)+1/11]
=12×(1+1/2+1/3+...+1/10) -6 -6/11+12 -12×(1+1/2+1/3+...+1/10)
=12-6 -6/11
=60/11

Answer 2

是希望杯冲刺班的题吧？

Answer 3

后面那个分母是除还是加？

追问

是相加