Question

用Sn-Sn-1求an通项公式……一题!？

Answer 1

a_(n+1) = S_(n+1) - S_n
= [(n + 1)^2 + 0.5 a_(n+1)] - [n^2 + 0.5 a_n]
= (n + 1)^2 - n^2 + 0.5 a_(n+1) - 0.5 a_n
= 2 n + 1 + 0.5 a_(n+1) - 0.5 a_n ,
所以 a_(n+1) = - a_n + 4 n + 2 ,
所以 a_n = - a_(n-1) + 4 n - 2 = - [- a_(n-2) + 4 n - 6] + 4 n - 2
= a_(n-2) + 4 ,
又 S_1 = a_1 = 1 + 0.5 a_1 ,S_2 = a_1 + a_2 = 4 + 0.5 a_2 ,
所以 a_1 = 2 ,a_2 = 4 ,
所以 {a_(2n-1)} 是以 2 为首项,以 4 为公差的等差数列,
{a_(2n)} 是以 4 为首项,以 4 为公差的等差数列,
所以 a_(2n-1) = 2 + 4 (n - 1) = 4 n - 2 ,
a_(2n) = 4 + 4 (n - 1) = 4 n ,
综上,a_n = 2 n .,8,Sn-Sn-1=2n+0.5(an-an-1)=an
0.5an=2n-0.5an-1
an=4n-an-1
an+an-1=4n
an-1+an-2=4(n-1)
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a2-a1=4*2
左右分别相加
an-a1=4n+4(n-1)+...+4*2=4*(2+n)*(n-1)/2=2(n+2)(n-1)
S1=a1=1+0.5=1.5
an=2(n+2)(n-1)+1.5,0,用Sn-Sn-1求an通项公式……一题!
已知数列{an}的前n项和为Sn,且对一切正整数n都有Sn=n^2+0.5an,求数列{an}通项公式.