已知数列{an}中,a1=2,an+1=an²+2an(n∈N+)
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已知数列{an}中,a1=2,an+1=an²+2an(n∈N+)
(1)证明数列{log2(an+1)}是等比数列,并求数列an的通项公式
(2)记数列{bn}满足bn=(an)+1/(an+1),求证bn=(an+1)-(an)/(an)(an+1),并求数列的前n项和Sn
(1)证明数列{log2(an+1)}是等比数列,并求数列an的通项公式
(2)记数列{bn}满足bn=(an)+1/(an+1),求证bn=(an+1)-(an)/(an)(an+1),并求数列的前n项和Sn
a(n+1)=an^2+2an
a(n+1)+1=(an+1)^2
log2[(a(n+1)+1]=2log2[(an)+1]
log2[(a(n+1)+1]/log2[an+1]=2
{log2[a(n+1)+1]}等比数列,公比:2,首项:1
log2[(an)+1]=2^(n-1)
an=2^[2^(n-1)]-1=4^(n-1)-1
a(n+1)=an^2+2an
bn=[(an)+1]/(an+1),(分子分母同乘an)
=>bn=[(an+1)-(an)]/(an)(an+1)=1/an-1/a(n+1)
a(n+1)+1=(an+1)^2
log2[(a(n+1)+1]=2log2[(an)+1]
log2[(a(n+1)+1]/log2[an+1]=2
{log2[a(n+1)+1]}等比数列,公比:2,首项:1
log2[(an)+1]=2^(n-1)
an=2^[2^(n-1)]-1=4^(n-1)-1
a(n+1)=an^2+2an
bn=[(an)+1]/(an+1),(分子分母同乘an)
=>bn=[(an+1)-(an)]/(an)(an+1)=1/an-1/a(n+1)
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