Question

已知数列｛an｝的通项公式是an=5/2^n-7/（n^2+3n+2）求an各项和的极限

Answer 1

an=5/2^n-7/（n^2+3n+2）=5/2^n-7/(n+1)(n+2)=5*(1/2)^n-7*[1/(n+1)-1/(n+2)]

前n项和等于

=5*【1/2+1/2^2+1/2^3+……1/2^n】-7*【1/2-1/3+1/3-1/4+1/4-1/5……+1/(n+1)-1/(n+2)】

=5*【1-1/2^n】-7*【1/2-1/(n+2)】

极限就是

=5*1-7/2

=1.5