已知数列{an}的前n和为Sn,且Sn=2an+n^2－3n－2 n为正整数求证：1数列是等比数列2设bn=an*cos

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已知数列{an}的前n和为Sn,且Sn=2an+n^2－3n－2 n为正整数求证：1数列是等比数列2设bn=an*cosn180度,求数
列bn的前n项和p

s1=a1=2a1+1^2-3*1-2
a1=2a1-4
a1=4
sn=2an+n^2-3n-2
s(n-1)=2a(n-1)+(n-1)^2-3(n-1)-2
sn-s(n-1)=2an+n^2-3n-2-2a(n-1)-(n-1)^2+3(n-1)+2
an=2an-2a(n-1)+n^2-(n-1)^2-3n+3(n-1)
an=2an-2a(n-1)+2n-2
an=2a(n-1)-2n+2
an-2n=2a(n-1)-2(n-1)
(an-2n)/[a(n-1)-2(n-1)]=2
所以an-2n是以2为公比的等比数列
an-2n=(a1-2*1)*2^(n-1)
an-2n=(4-2)*2^n
an-2n=2^n
an=2^n+2n
cosn180度=(-1)^n
bn=ancosn180度
=(-1)^n*(2^n+2n)
当n为偶数时,设n=2k（k∈N*）
Pn=b1+b2+…+bn=[b1+b3+…+b(2k-1)]+(b2+b4+…+b2k)
={-(2+2×1)-(2^3+2×3)-…-[2^(2k-1)+2(2k-1)]}+[(2^2+2×2)+(2^4+2×4)+…+(2^2k+2k)]
=-[(2+2^3+…+2^(2k-1)]-2[1+3+…+(2k-1)]+(2^2+2^4+…+2^2k）+2(2+4+…+2k)
=-[2-2^2+2^3-2^4+…+2^(2k-1)-2^2k]+2[-1+2-3+4-…-(2k-1)+2k]
=-2[1-(-2)^2k]/[1-(-2)]+2k
=-2[1-2^2k]/[1-(-2)]+2k
=-2[1-2^2k]/3+2k
=-2(1-2^n)/3+n
=2(2^n-1)/3+n
当n为奇数时,设n=2k-1（k∈N*）,同理可得
Pn=-{2^[(2k-1)+1]+2}/3-[(2k-1)+1]
=-[2^(n+1)+2]/3-(n+1)
=-2^(n+1)/3-2/3-n-1
=-2^(n+1)/3-n-5/3
综上所述,
Pn=2(2^n-1)/3+n (n为偶数)
Pn=-2^(n+1)/3-n-5/3 (n为奇数)

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