若数列AN的通项公式为An=1⼀(n√(n+1)+(n+1)√n)问An的前99项和为?求教,要详细过程.谢谢!

an=1/[n√(n+1)+(n+1)√n]
=[1/√n(n+1)][1/[√n +√(n+1)]
=[1/√n(n+1)][√(n+1)-√n] 这一步是后面一项分母有理化。
=√(n+1)/√n(n+1)-√n/√n(n+1)
=1/√n -1/√(n+1)
a1+a2+...+a99
=1/√1-1/√2+1/√2-1/√3+...+1/√99-1/√100
=1/√1-1/√100
=1-1/10
=9/10

An=1/[n√(n+1)+(n+1)√n]
=[(n+1)√n-n√(n+1)]/[(n+1)²n-n²(n+1)]
=[(n+1)√n-n√(n+1)]/[n³+2n²+n-n³-n²]
=[(n+1)√n-n√(n+1)]/(n²+n)
=[(n+1)√n-n√(n+1)]/[n(n+1)]
=[(n+1)√n]/[n(n+1)]－[n√(n+1)]/[n(n+1)]
=1/√n－1/√(n+1)
所以S99=1/1－1/√2＋1/√2－1/√3+……+1/√99－1/√100
=1－1/√100
=9/10

答案是9/10.
等我写过程给你……

分母提公因式然后并向裂项相消

an=1/[n√(n+1)+(n+1)√n]
=1/[√n√(n+1)(√n+√(n+1)]
=[√n-√(n+1)]/[√n√(n+1)(√n+√(n+1)][√n-√(n+1)]
=[√n-√(n+1)]/[√n√(n+1)(n-(n+1)]
=-[√n-√(n+1)]/[√n√(n+1)]
=[√(n+1)-√n]/[√n√(n+1)]
=1/√n-1/√(n+1)

s99
=1/√1-1/√2+1/√2-1/√3+1/√3-1/√4+.............+1/√99-1/√100
=1-1/√100
=1-1/10
=9/10

An=1/{sqrt(n*(n+1))*[sqrt(n)+sqrt(n+1)]}
=[1/sqrt(n) - 1/sqrt(n+1)]* 1/{[sqrt(n+1)+sqrt(n)]*[sqrt(n+1)-sqrt(n)]}
=1/sqrt(n) - 1/sqrt(n+1);
so A1+...+A99=1-1/sqrt(100)=9/10 .