精英家教網(wǎng) > 高中數(shù)學(xué) > 題目詳情
數(shù)列{an}中,已知a1=1,n≥2時(shí),an=
an-1+
-
.?dāng)?shù)列{bn}滿足:bn=3n-1(an+1).
(1)求證:數(shù)列{bn}是等差數(shù)列;
(2)求數(shù)列{an}的前n項(xiàng)和Sn.
1 |
3 |
2 |
3n-1 |
2 |
3 |
(1)求證:數(shù)列{bn}是等差數(shù)列;
(2)求數(shù)列{an}的前n項(xiàng)和Sn.
考點(diǎn):數(shù)列的求和,等差關(guān)系的確定
專題:等差數(shù)列與等比數(shù)列
分析:(1)根據(jù)題中的遞推關(guān)系式進(jìn)行恒等變換求出數(shù)列是等差數(shù)列.
(2)利用(1)的結(jié)論,進(jìn)一步利用所構(gòu)造的新數(shù)列利用乘公比錯(cuò)位相減法求數(shù)列的和.
(2)利用(1)的結(jié)論,進(jìn)一步利用所構(gòu)造的新數(shù)列利用乘公比錯(cuò)位相減法求數(shù)列的和.
解答:
(1)證明:由an=
an-1+
-
得:3n-1an=3n-2an-1+2-2×3n-2
∴3n-1(an+1)=3n-1an+3n-1=3n-2an-1+2+3n-2=3n-2(an-1+1)+2
即bn=bn-1+2⇒bn-bn-1=2(n≥2)
又b1=31-1(a1+1)=2
∴數(shù)列{bn}是首項(xiàng)為2,公差為2的等差數(shù)列.
(2)解:由(1)知,bn=2+(n-1)×2=2n,
∴3n-1(an+1)=2n⇒an=
-1
記Tn=
+
+
+…+
,
則
Tn=
+
+…+
+
兩式相減得:
Tn=2(1+
+
+…+
)-
=
-
=3-
∴Tn=
-
因此,Sn=Tn-n=
-
-n
1 |
3 |
2 |
3n-1 |
2 |
3 |
得:3n-1an=3n-2an-1+2-2×3n-2
∴3n-1(an+1)=3n-1an+3n-1=3n-2an-1+2+3n-2=3n-2(an-1+1)+2
即bn=bn-1+2⇒bn-bn-1=2(n≥2)
又b1=31-1(a1+1)=2
∴數(shù)列{bn}是首項(xiàng)為2,公差為2的等差數(shù)列.
(2)解:由(1)知,bn=2+(n-1)×2=2n,
∴3n-1(an+1)=2n⇒an=
2n |
3n-1 |
記Tn=
2 |
1 |
4 |
3 |
6 |
32 |
2n |
3n-1 |
則
1 |
3 |
2 |
3 |
4 |
32 |
2(n-1) |
3n-1 |
2n |
3n |
兩式相減得:
2 |
3 |
1 |
3 |
1 |
32 |
1 |
3n-1 |
2n |
3n |
=
2[1-(
| ||
1-
|
2n |
3n |
2n+3 |
3n |
∴Tn=
9 |
2 |
2n+3 |
2×3n-1 |
因此,Sn=Tn-n=
9 |
2 |
2n+3 |
2×3n-1 |
點(diǎn)評:本題考查的知識要點(diǎn):利用遞推關(guān)系式求出數(shù)列是等差數(shù)列,并求出數(shù)列的通項(xiàng)公式乘公比錯(cuò)位相減法的應(yīng)用.屬于基礎(chǔ)題型.