分析:(Ⅰ) 當(dāng)n=1時(shí),(t-1)S
1+(2t+1)a
1=t,可求的首項(xiàng)a
1=
;當(dāng)n≥2時(shí),(t-1)S
n+(2t+1)a
n=t,(t-1)S
n-1+(2t+1)a
n-1=t,兩式相減可得(t-1)a
n+(2t+1)a
n-(2t+1)a
n-1=0,從而有
=,故可知數(shù)列{a
n}是以
為公比,
為首項(xiàng)的等比數(shù)列;
(II)由(Ⅰ)可知,
f(t)=(t>0),
bn+1=3f(),則b
n+1=b
n+2,從而可得數(shù)列{b
n}是以2為公差,首項(xiàng)為1的等差數(shù)列,從而b
n=2n-1由于涉及(-1)
n+1,故分n為偶數(shù)及奇數(shù)分類求和.
解答:證明:(Ⅰ) 當(dāng)n=1時(shí),(t-1)S
1+(2t+1)a
1=t,∴a
1=
當(dāng)n≥2時(shí),(t-1)S
n+(2t+1)a
n=t,(t-1)S
n-1+(2t+1)a
n-1=t
∴(t-1)a
n+(2t+1)a
n-(2t+1)a
n-1=0
∴3ta
n=(2t+1)a
n-1,t>0
∴
=,a1=∴數(shù)列{a
n}是以
為公比,
為首項(xiàng)的等比數(shù)列;
解:(II)由(Ⅰ)可知,
f(t)=(t>0),
bn+1=3f(),則b
n+1=b
n+2
所以,數(shù)列{b
n}是以2為公差,首項(xiàng)為1的等差數(shù)列
即b
n=2n-1
①當(dāng)n為奇數(shù)時(shí),
b
1b
2-b
2b
3+b
3b
4-b
4b
5+…+(-1)
n+1b
nb
n+1=b
1b
2+b
3(b
4-b
2)+b
5(b
6-b
4)+…+b
n(b
n+1-b
n-1)
=3+4(b
3+b
5+…+b
n)
=2n
2+2n-1
②當(dāng)n為偶數(shù)時(shí),
b
1b
2-b
2b
3+b
3b
4-b
4b
5+…+(-1)
n+1b
nb
n+1=b
2(b
1-b
3)+b
4(b
3-b
5)+…+b
n(b
n-1-b
n+1)
=-4(b
2+b
4+…+b
n)
=-(2n
2+2n)
所以,原式=
| 2n2+2n-1 n為奇數(shù) | -(2n2+2n) n為偶數(shù) |
| |
點(diǎn)評:本題以數(shù)列遞推式為載體,考查等比數(shù)列的定義,考查等差數(shù)列的通項(xiàng),同時(shí)考查了分類討論的數(shù)學(xué)思想,綜合性強(qiáng).