Question

初二数学二次根式题

计算:1/(2+√2)+1/(3√2+3√3)+1/(4√3+3√4)+…+1/(100√99+99√100)

Answer 1

1/[n√(n+1)+(n+1)√n]
=[n√(n+1)-(n+1)√n]/[n√(n+1)+(n+1)√n][n√(n+1)-(n+1)√n]
=[n√(n+1)-(n+1)√n]/[n^2(n+1)-n(n+1)^2]
=[n√(n+1)-(n+1)√n]/[-n(n+1)]
=(n+1)√n/[n(n+1)]-n√(n+1)/[n(n+1)]
=1/√n-1/√(n+1)

所以原式=(1/√1-1/√2)+(1/√2-1/√3)+……+(1/√99-1/√100)
=1/√1-1/√100
=1-1/10
=9/10

Answer 2

所有分母有理化！立刻就可以看出来规律！