Question

化简 1) √[(1998*1999*2000*2001+1)/4] 2) 1/(2√1+1√2)+1/(3√2+2√3)+……+1/(100√99+99√100)

化简 1) √[(1998*1999*2000*2001+1)/4]
2) 1/(2√1+1√2)+1/(3√2+2√3)+……+1/(100√99+99√100)

Answer 1

1、
令a=1998
则1998*1999*2000*2001+1
=a(a+1)(a+2)(a+3)+1
=[a(a+3)][(a+1)(a+2)]+1
=(a²+3a)[(a²+3a)+2]+1
=(a²+3a)²+2(a²+3a)+1
=(a²+3a+1)²
所以原式=√[(a²+3a+1)²/4]
=(a²+3a+1)/2
=(1998²+3×1998+1)/2
=3997999/2

2、
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√(n+1)]/[n(n+1)²-n²(n+1)]
=[(n+1)√n-n√(n+1)]/n(n+1)
=(n+1)√n/n(n+1)-n√(n+1)/n(n+1)
=√n/n-√(n+1)/(n+1)

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