An=1/(n平方+4n) 求An的前n项和
1个回答

an=1/(n^2+4n)

=1/[n(n+4)]

=1/4*[1/n-1/(n+4)]

∴前n项和

=a1+a2+a3+a4+a5...+a(n-4)+a(n-3)+a(n-2)+a(n-1)+an

=1/4*(1-1/5+1/2-1/6+1/3-1/7+1/4-1/8+1/5-1/9+1/6-1/10+

.1/(n-4)-1/n+1/(n-3)-1/(n+1)+1/(n-2)-1/(n+2)+1/(n-1)-1/(n+3)+1/n-1/(n+4)]

=1/4[1+1/2+1/3+1/4-1/(n+1)-1/(n+2)-1/(n+3)-1/(n+4)]

就是裂项求和,第5项开始能消去

结果就是这个,顶多通分