计算1/1*2 + 1/2*3 1/3*4 +1/4*5 +1/5*6计算1/1*3 +1/3*5 +1/5*7 +1/7*9 +1/9*11化简1/x(x+3) + 1/(x+3)(x+6)+ 1/(x+6)(x+9) +1/(x+9)(x+12)
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计算1/1*2 + 1/2*3 1/3*4 +1/4*5 +1/5*6计算1/1*3 +1/3*5 +1/5*7 +1/7*9 +1/9*11化简1/x(x+3) + 1/(x+3)(x+6)+ 1/(x+6)(x+9) +1/(x+9)(x+12)
计算1/1*2 + 1/2*3 1/3*4 +1/4*5 +1/5*6
计算1/1*3 +1/3*5 +1/5*7 +1/7*9 +1/9*11
化简1/x(x+3) + 1/(x+3)(x+6)+ 1/(x+6)(x+9) +1/(x+9)(x+12)
计算1/1*2 + 1/2*3 1/3*4 +1/4*5 +1/5*6计算1/1*3 +1/3*5 +1/5*7 +1/7*9 +1/9*11化简1/x(x+3) + 1/(x+3)(x+6)+ 1/(x+6)(x+9) +1/(x+9)(x+12)
1/(1×2) + 1/(2×3)+ 1/(3×4) +1/(4×5) +1/(5×6)
做这题之前请看下面 一些式子:
1/(1×2)=1-(1/2)=1/2
1/(2×3)=(1/2)-(1/3)=1/6
1/(3×4)=(1/3)-(1/4)=1/12
…………
1/(1×2) + 1/(2×3)+ 1/(3×4) +1/(4×5) +1/(5×6)
原式=1-(1/2)+(1/2)-(1/3)+(1/3)-(1/4)+(1/4)-(1/5)+(1/5)-(1/6)
=1-(1/6)
=5/6
1/(1×3) +1/(3×5 )+1/(5×7) +1/(7×9) +1/(9×11)
1/3*5=0.5×[(1/3)-(1/5)]
1/5*7=0.5×[(1/5-1/7)] ……………
∴ 1/(1×3) +1/(3×5 )+1/(5×7) +1/(7×9) +1/(9×11)
原式=0.5[1-(1/3)+(1/3)-(1/5)+(1/5)-……-(1/9)+(1/9)-(1/11)]
=0.5×[1-(1/11)]
=0.5×(10/11)
=5/11
1/x(x+3) + 1/(x+3)(x+6)+ 1/(x+6)(x+9) +1/(x+9)(x+12)
1/x(x+3)=1/3{(1/x)-[1/(x+3)]}
1/(x+3)(x+6)=1/3{[1/(x+3)]-1/(x+6)}…………
1/x(x+3) + 1/(x+3)(x+6)+ 1/(x+6)(x+9) +1/(x+9)(x+12)
原式=1/3{(1/x)-[1/(x+3)]+[1/(x+3)]-[1/(x+6)]+[1/(x+6)]-………-[1/(x+9)]+[1/(x+9)]-[1/(x+11)]}
=1/3{(1/x)-[1/(x+11)]}
=1/3[(x+10)/(x²+11x)]
=(x+10)/(3x²+33x)
第一式的每一项可以拆成这样
例如1/3*5=0.5*(1/3-1/5)
1/5*7=0.5*(1/5-1/7)
这样前后项相消,得 0.5*(1-1/11)=5/11
同理第二个式子的通项可以拆成这样
1/x(x+3)=(1/3)*(1/x-1/(x+3))
也是前后项相消,得 (1/3)*(1/x-1/(x+12))
1/(1×2) + 1/(2×3)+ 1/(3×4) +1/(4×5) +1/(5×6) 是不是这样的???