0.5*(1/k)*[1/(k+2)]=0.25*[1/k-1/(k+2)]

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0.5*(1/k)*[1/(k+2)]=0.25*[1/k-1/(k+2)]0.5*(1/k)*[1/(k+2)]=0.25*[1/k-1/(k+2)]0.5*(1/k)*[1/(k+2)]=0.25

0.5*(1/k)*[1/(k+2)]=0.25*[1/k-1/(k+2)]
0.5*(1/k)*[1/(k+2)]=0.25*[1/k-1/(k+2)]

0.5*(1/k)*[1/(k+2)]=0.25*[1/k-1/(k+2)]
两边同时乘以k*(k+2)
0.5=0.25(k+2-k)
此式恒成立啊!