1/(2倍根号1+根号2)+1/(3倍根号2+2倍根号3)+1/(4倍根号3+3倍根号4)+···+1/(100倍根号9
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1/(2倍根号1+根号2)+1/(3倍根号2+2倍根号3)+1/(4倍根号3+3倍根号4)+···+1/(100倍根号99+99倍根号10
考察一般项:
an=1/[(n+1)√n+n√(n+1)]
=[(n+1)√n-n√(n+1)]/[n(n+1)^2-(n+1)n^2]
=[(n+1)√n-n√(n+1)]/(n^3+2n^2+n-n^3-n^2)
=[(n+1)√n-n√(n+1)]/[n(n+1)]
=1/√n-1/√(n+1)
1/(2√1+√2)+1/(3√2+2√3)+...+1/(100√99+99√100)
=1-1/√2+1/√2-1/√3+1/√3-1/√4+...+1/√98-1/√99+1/√99-1/√100
=1-1/√100
=1-1/10
=9/10
an=1/[(n+1)√n+n√(n+1)]
=[(n+1)√n-n√(n+1)]/[n(n+1)^2-(n+1)n^2]
=[(n+1)√n-n√(n+1)]/(n^3+2n^2+n-n^3-n^2)
=[(n+1)√n-n√(n+1)]/[n(n+1)]
=1/√n-1/√(n+1)
1/(2√1+√2)+1/(3√2+2√3)+...+1/(100√99+99√100)
=1-1/√2+1/√2-1/√3+1/√3-1/√4+...+1/√98-1/√99+1/√99-1/√100
=1-1/√100
=1-1/10
=9/10
1/(2倍根号1+根号2)+1/(3倍根号2+2倍根号3)+1/(4倍根号3+3倍根号4)+···+1/(100倍根号9
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