√(1+1/1²+1/2²)+√(1+1/2²+1/3²)+…+√﹙1+1/20
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√(1+1/1²+1/2²)+√(1+1/2²+1/3²)+…+√﹙1+1/2003²+1/2004²)=
∵1+1/n^2+1/(n+1)^2
={[n(n+1)]^2+(n+1)^2+n^2}/[n(n+1)]^2
={[n(n+1)]^2+[(n+1)^2-2n(n+1)+n^2]+2n(n+1)}/[n(n+1)]^2
={[n(n+1)]^2+2n(n+1)+[(n+1)-n]^2}/[n(n+1)]^2
={[n(n+1)]^2+2n(n+1)+1}/[n(n+1)]^2
=[n(n+1)+1]^2/[n(n+1)]^2,
∴√[1+1/n^2+1/(n+1)^2]
=[n(n+1)+1]/[n(n+1)]=1+1/[n(n+1)]=1+1/n-1/(n+1).
在上式中,依次令n=1、2、3、4、······、2003、2004,然后相加,得:
√(1+1/1^2+1/2^2)+√(1+1/2^2+1/3^2)+·····+√(1+1/2003^2+1/2004^2)
=(1+1/1-1/2)+(1+1/2-1/3)+·····+(1+1/2003-1/2004)
=1+1/1-1/2004
=(4008-1)/2004
=4007/2004.
={[n(n+1)]^2+(n+1)^2+n^2}/[n(n+1)]^2
={[n(n+1)]^2+[(n+1)^2-2n(n+1)+n^2]+2n(n+1)}/[n(n+1)]^2
={[n(n+1)]^2+2n(n+1)+[(n+1)-n]^2}/[n(n+1)]^2
={[n(n+1)]^2+2n(n+1)+1}/[n(n+1)]^2
=[n(n+1)+1]^2/[n(n+1)]^2,
∴√[1+1/n^2+1/(n+1)^2]
=[n(n+1)+1]/[n(n+1)]=1+1/[n(n+1)]=1+1/n-1/(n+1).
在上式中,依次令n=1、2、3、4、······、2003、2004,然后相加,得:
√(1+1/1^2+1/2^2)+√(1+1/2^2+1/3^2)+·····+√(1+1/2003^2+1/2004^2)
=(1+1/1-1/2)+(1+1/2-1/3)+·····+(1+1/2003-1/2004)
=1+1/1-1/2004
=(4008-1)/2004
=4007/2004.
√(1+1/1²+1/2²)+√(1+1/2²+1/3²)+…+√﹙1+1/20
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