考點:等比數(shù)列的前n項和,等比關(guān)系的確定
專題:等差數(shù)列與等比數(shù)列
分析:(1)易得數(shù)列{a
n}也是等比數(shù)列,由題意易得公比q=±2,分別可得通項公式;
(2)當(dāng)q=2時,S
n+λ=2•2
n-1+λ-1,當(dāng)且僅當(dāng)λ-1=0即λ=1時為等比數(shù)列;同理當(dāng)q=-2時可得λ=-
,綜合可得.
解答:
解:(1)∵數(shù)列{
}是等比數(shù)列,∴數(shù)列{a
n}也是等比數(shù)列,
設(shè)數(shù)列{a
n}的公比為q,則a
33=a
2a
3a
4=64,解得a
3=4,
∴q
2=
=4,解得q=±2,
當(dāng)q=2時,a
n=2
n-1;當(dāng)q=-2時,a
n=(-2)
n-1;
(2)當(dāng)q=2時,S
n+λ=
+λ=2•2
n-1+λ-1,
當(dāng)且僅當(dāng)λ-1=0即λ=1時數(shù)列{S
n+λ}是2為首項2為公比的等比數(shù)列;
同理當(dāng)q=-2時,S
n+λ=
+λ=
•(-2)
n-1+λ+
,
當(dāng)且僅當(dāng)λ+
=0即λ=-
時數(shù)列{S
n+λ}是
為首項-2為公比的等比數(shù)列,
∴λ的值為1或
-.
點評:本題考查等比數(shù)列的求和公式和性質(zhì),涉及分類討論的思想,屬中檔題.