Question

在数列an中.a1等于a,an+1等于Sn+2n次方.设bn=Sn除以2（n-1）次方,球bn通项公式

Answer 1

a(n+1)=Sn+2ⁿ

S(n+1)-Sn=Sn+2ⁿ

S(n+1)-2Sn=2ⁿ

等式两边同除以2ⁿ

S(n+1)/2ⁿ-Sn/2^(n-1) =1,为定值.

S1/2^0=a1/1=a

数列{Sn/2^(n-1)}是以a为首项,1为公差的等差数列.

bn=Sn/2^(n-1),数列{bn}是以a为首项,1为公差的等差数列.

bn=a+1×(n-1)=n +a-1

数列{bn}的通项公式为bn=n+a-1.