tính lim ( n^2 +n - tổng xích ma từ k-1 đến n ( 2k^3 +8 k^2 +6k -1 )/( k^2 +4k +3)

Đáp án D

Ta có

2k3+8k2+6k−1k2+4k+3=2kk+1k+3k+1k+3−1k+1k+3=2k−121k+1−1k+3

⇒∑k=1n2k3+8k2+6k−1k2+4k+3=∑k=1n2k−121k+1−1k+3

=2.1+2+...+n−1211+1−11+3+...+1n−1+1−1n−1+3+1n+1−1n+3

=2nn+12−1212+13−1n+2−1n+3=nn+1−1256−2n+5n+2n+3

suy ra 

limn2+n−∑k=1n2k3+8k2+6k−1k2+4k+3=limn2+n−n2+n−512+2n+52n+2n+3

=lim512−2n+52n2+10n+12=lim512−lim2n+5n22+10n+12n2=512