设等差数列{an}的前n项和为Sn,且Sn=(an+12)2,(n∈N*),若bn=(?1)nSn,求数列{bn}的前n项和Tn
设等差数列{an}的前n项和为Sn,且Sn=(an+12)2,(n∈N*),若bn=(?1)nSn,求数列{bn}的前n项和Tn....
设等差数列{an}的前n项和为Sn,且Sn=(an+12)2,(n∈N*),若bn=(?1)nSn,求数列{bn}的前n项和Tn.
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因为a1=S1=(
)2,所以 a1=1.
设公差为d,则有a1+a2=2+d=S2=(
)2.
解得d=2或d=-2(舍).
所以an=2n-1,Sn=n2.
所以 bn=(?1)n?n2.
(1)当n为偶数时,Tn=?12+22?32+42?…+(?1)nn2
=(22-12)+(42-32)+…+[n2-(n-1)2]
=3+7+11+…+(2n?1)=
;
(2)当n为奇数时,Tn=Tn?1?n2
=
?n2=?
=?
.
综上,Tn=(?1)n?
.
a1+1 |
2 |
设公差为d,则有a1+a2=2+d=S2=(
2+d |
2 |
解得d=2或d=-2(舍).
所以an=2n-1,Sn=n2.
所以 bn=(?1)n?n2.
(1)当n为偶数时,Tn=?12+22?32+42?…+(?1)nn2
=(22-12)+(42-32)+…+[n2-(n-1)2]
=3+7+11+…+(2n?1)=
n(n+1) |
2 |
(2)当n为奇数时,Tn=Tn?1?n2
=
(n?1)?n |
2 |
n2+n |
2 |
n(n+1) |
2 |
综上,Tn=(?1)n?
n(n+1) |
2 |
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