第3个回答 2021-07-15
(4)
1 +1/2 +1/2^2+...+1/2^n = 2[1 -(1/2)^(n+1)]
(n+1)+2(n+2)+...+n(n+n)
=[n+2n+....+n(n)] +(1^2+2^2+...+n^2)
=n(1+2+...+n) + (1/6)n(n+1)(2n+1)
=(1/2)n^2.(n+1) + (1/6)n(n+1)(2n+1)
n^3 的系数 = 1/2 + 1/3 = 5/6
lim(n->无穷) n^3. (1 +1/2 +1/2^2+...+1/2^n) /[ (n+1)+2(n+2)+...+n(n+n) ]
=lim(n->无穷) n^3. { 2[1 -(1/2)^(n+1)] } / [(5/6)n^3 ]
=lim(n->无穷) { 2[1 -(1/2)^(n+1)] } / (5/6)
=12/5