Re: [中學] 根號
: 求解，謝謝各位：）
[1+(1/1)^2+(1/2)^2]^(1/2)=[1+1+1/4]^(1/2) = 3/ 2=( 2+1)/(1*2)
[1+(1/2)^2+(1/3)^2]^(1/2)=[1+1/4+1/9]^(1/2) = 7/ 6=( 6+1)/(2*3)
[1+(1/3)^2+(1/4)^2]^(1/2)=[1+1/9+1/16]^(1/2) =13/12=(12+1)/(3*4)
[1+(1/4)^2+(1/5)^2]^(1/2)=[1+1/16+1/25]^(1/2)=21/20=(20+1)/(4*5)
可歸納得Σ(k=1,199) (k(k+1)+1)/(k(k+1))
=Σ(k=1,199) 1+1/(k(k+1))
=Σ(k=1,199) 1+Σ(k=1,199) 1/(k(k+1))
=199*1+Σ(k=1,199) 1/k-1/(k+1)
=199+1-1/2+1/2-1/3+1/3-1/4+...-1/199+1/199-1/200
=199+1-1/200=199+199/200(199又200分之199)
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06/20 23:38, , 1F
06/20 23:38, 1F
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