已知数列{an}的前n项和为sn，an=1/(√(n-1)+√n)(√(n-1)+√(n+1))(√n+√(n+1)),
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1/[(n-1)^(1/2)+(n+1)^(1/2)] = [(n+1)^(1/2) - (n-1)^(1/2)]/[(n+1)-(n-1)] = [(n+1)^(1/2)-(n-1)^(1/2)]/2,
(1/2)/[(n-1)^(1/2)+n^(1/2)] - (1/2)/[n^(1/2)+(n+1)^(1/2)] = {[(n+1)^(1/2) - (n-1)^(1/2)]/2}/{[(n-1)^(1/2) + n^(1/2)][n^(1/2)+(n+1)^(1/2)]}
= 1/{[(n-1)^(1/2) + n^(1/2)][(n-1)^(1/2) + (n+1)^(1/2)][n^(1/2)+(n+1)^(1/2)]} = a(n),
s(n)=(1/2){1/[0+1] - 1/[1+2^(1/2)] + 1/[2^(1/2)+3^(1/2)] - 1/[1+2^(1/2)] + ... + 1/[(n-2)^(1/2) + (n-1)^(1/2) ] - 1/[(n-1)^(1/2) + n^(1/2)] + 1/[(n-1)^(1/2) + n^(1/2)] - 1/[n^(1/2) + (n+1)^(1/2)]}
=(1/2){1 - 1/[n^(1/2)+(n+1)^(1/2)]}
=(1/2){1- [(n+1)^(1/2) - n^(1/2)]/[n+1-n]}
=(1/2){1-(n+1)^(1/2) + n^(1/2)}
=[1+n^(1/2) - (n+1)^(1/2)]/2
(1/2)/[(n-1)^(1/2)+n^(1/2)] - (1/2)/[n^(1/2)+(n+1)^(1/2)] = {[(n+1)^(1/2) - (n-1)^(1/2)]/2}/{[(n-1)^(1/2) + n^(1/2)][n^(1/2)+(n+1)^(1/2)]}
= 1/{[(n-1)^(1/2) + n^(1/2)][(n-1)^(1/2) + (n+1)^(1/2)][n^(1/2)+(n+1)^(1/2)]} = a(n),
s(n)=(1/2){1/[0+1] - 1/[1+2^(1/2)] + 1/[2^(1/2)+3^(1/2)] - 1/[1+2^(1/2)] + ... + 1/[(n-2)^(1/2) + (n-1)^(1/2) ] - 1/[(n-1)^(1/2) + n^(1/2)] + 1/[(n-1)^(1/2) + n^(1/2)] - 1/[n^(1/2) + (n+1)^(1/2)]}
=(1/2){1 - 1/[n^(1/2)+(n+1)^(1/2)]}
=(1/2){1- [(n+1)^(1/2) - n^(1/2)]/[n+1-n]}
=(1/2){1-(n+1)^(1/2) + n^(1/2)}
=[1+n^(1/2) - (n+1)^(1/2)]/2
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an=1/(√(n-1)+√n)(√(n-1)+√(n+1))(√n+√(n+1)),=[1/(√(n-1)+√n)-1/(√n+√(n+1)]/2
a1=[1-1/(1+√2)]/2，a2=[1/(1+√2)-1/(1/(√2+√3)]/2,┄┄[1/(√(n-1)+√n)-1/(√n+√(n+1)]/2
Sn=[1-1/(√n+√(n+1)]/2=[1+√n-√(n+1)]/2。
a1=[1-1/(1+√2)]/2，a2=[1/(1+√2)-1/(1/(√2+√3)]/2,┄┄[1/(√(n-1)+√n)-1/(√n+√(n+1)]/2
Sn=[1-1/(√n+√(n+1)]/2=[1+√n-√(n+1)]/2。
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没打完就发上来了啊
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