Question

设各项均为正数的数列an的前n项和为sn满足sn=1/4(an+1)

已知各项均为正数的数列{an}的前n项和为Sn,根号Sn是1/4与(an+1)^2的等比中项.
(1)求证:数列{an}是等差数列
(2)若bn=(an)/(2^n),数列bn的前n项和为Tn,求Tn

Answer 1

n>=2时,S[n]=1/4 * (a[n]+1)^2; S[n-1]=1/4 * (a[n-1]+1)^2
两式相减得到a[n]=1/4 * (a[n]^2+2a[n]-a[n-1]^2-2a[n-1])
化简得到a[n]^2-a[n-1]^2=2a[n]+2a[n-1]
得到a[n]-a[n-1]=2所以是等差数列.首项是1,公差是2
a[n]=2n-1
第二步不难,但写出来比较麻烦
答案是Tn=3-（2n+3)/(2^n)