已知:数列an的前n项和为Sn=2n^2-6n+1（n∈Z+）
已知:数列an的前n项和为Sn=2n^2-6n+1（n∈Z+）
若bn=|an|,求数列{bn}的前n项和
若bn=|an|,求数列{bn}的前n项和
赞 jpe林 幼苗
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Sn=2n^2-6n+1
n=1, a1=-3
for n>=2
an =Sn -S(n-1)
= 2(2n-1) -6
=4n-8
ie
an=-3 ; n=1
=4n-8 ; n=2,3,4,...
bn =|an|
an >0
4n-8 >0
n > 2
b1= |a1|=3
b2=|a2| = 0
for n>=3
bn =4n-8
Tn =b1+b2+...+bn
n=1, 2
Tn =3
for n >=3
Tn = b1+b2+(b3+.+bn)
= 3+ (b3+b4+...+bn)
= 3+ (b3+bn)(n-3)/2
= 3 + ( 4n-4)(n-3)/2
= 3 + 2(n-1)(n-3)
= 2n^2 -8n + 9
n=1, a1=-3
for n>=2
an =Sn -S(n-1)
= 2(2n-1) -6
=4n-8
ie
an=-3 ; n=1
=4n-8 ; n=2,3,4,...
bn =|an|
an >0
4n-8 >0
n > 2
b1= |a1|=3
b2=|a2| = 0
for n>=3
bn =4n-8
Tn =b1+b2+...+bn
n=1, 2
Tn =3
for n >=3
Tn = b1+b2+(b3+.+bn)
= 3+ (b3+b4+...+bn)
= 3+ (b3+bn)(n-3)/2
= 3 + ( 4n-4)(n-3)/2
= 3 + 2(n-1)(n-3)
= 2n^2 -8n + 9
1年前
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