Question

高数题急求

第2 第4题

Answer 1

(2)
lim(n->∞) (1/n)[ 1/n+2/n)+...+(n-1)/n ]
=lim(n->∞) (1/n)[ 1/n+2/n)+...+(n-1)/n +n/n]
=∫(0->1) x dx
= (1/2)[x^2]|(0->1)
=1/2
(4)
2/log<x>4 + log<2>(x-1/4) +3=0
2(log<2>x/log<2>4)+ log<2>(x-1/4) +3=0
log<2>x +log<2>(x-1/4) =-3
log<2>[x.(x-1/4) ] = -3
x.(x-1/4) =1/8
2x.(4x-1) =1
8x^2-2x-1 =0
(4x+1)(2x-1)=0
x=1/2 or -1/4(rej)
ie
x=1/2
Sn=1+x+x^2+...+x^(n-1)
lim(n->∞) Sn
=lim(n->∞) (1-x^n)/(1-x)
=1/(1-1/2)
=2