Z treści zadania wiemy, że:
an=n−12n+1, n=1
a)
bn=2an+3=2⋅n−12n+1+3
Obliczamy granicę:
n→+∞limbn=n→+∞lim(2an+3)=
=n→+∞lim(2⋅n−12n+1+3)=
=n→+∞lim(2⋅n(1−n1)n(2+n1)+3)=
=2⋅1−02+0+3=2⋅2+3=4+3=7
b)
bn=21an−1=21⋅n−12n+1−1
Obliczamy granicę:
n→+∞limbn=n→+∞lim(21an−1)=
=n→+∞lim(21⋅n−12n+1−1)=
=n→+∞lim(21⋅n(1−n1)n(2+n1)−1)=
=21⋅1−02+0−1=21⋅2−1=1−1=0
c)
bn=1+an3=1+n−12n+13=1+3⋅2n+1n−1
Obliczamy granicę:
n→+∞limbn=n→+∞lim(1+an3)=
=n→+∞lim(1+3⋅2n+1n−1)=
=n→+∞lim(1+3⋅n(2+n1)n(1−n1))=
=1+3⋅2+01−0=1+3⋅21=1+23=25
d)
bn=an−3(an)2−6an+9=
=n−12n+1−3(n−12n+1)2−6⋅n−12n+1+9
Obliczamy granicę:
n→+∞limbn=n→+∞liman−3(an)2−6an+9=
=n→+∞limn−12n+1−3(n−12n+1)2−6⋅n−12n+1+9=
=n→+∞limn(1−n1)n(2+n1)−3(n(1−n1)n(2+n1))2−6⋅n(1−n1)n(2+n1)+9=
=1−02+0−3(1−02+0)2−6⋅1−02+0+9=
=2−34−6⋅2+9=
=−113−12=−1
