已知數(shù)列an的前n項(xiàng)和Sn滿足Sn-Sn-2=3(-1/2)^(n-1)(n>=3),且S1=1,S2=-3/2,求數(shù)列an

數(shù)學(xué)人氣：762 ℃時(shí)間：2019-10-23 11:26:48

優(yōu)質(zhì)解答

S1=1,S2=-3/2,所以
a1 = 1
a2 = -5/2
Sn - S = 3(-1/2)^(n-1)
S - S = 3(-1/2)^n
兩式相減
(S - Sn) - (S - S) = 3(-1/2)^n - 3(-1/2)^(n-1)
a - a = 9*(-1/2)^n
對(duì)于 n 為偶數(shù)情況
a3 - a1 = 9/2^2 = 9/4
a5 - a3 = 9/2^4 = 9/4^2
a7 - a5 = 9/2^6 = 9/4^3
……
a - a = 9/4^(n/2)
以上各等式相加
a - a1 = 9*[1/4 + 1/4^2 + …… + 1/4^(n/2)]
a - 1 = 9* (1/4)*[1 - 1/4^(n/2)]/(3/4) = 3*[1 - 1/4^(n/2)]
a = 4 - 3/4^(n/2) = 4 - 3/2^n
其中n為偶數(shù)
對(duì)于 n 為奇數(shù)數(shù)情況
a4 - a2 = -9/2^3
a6 - a4 = -9/2^5
a8 - a6 = -9/2^7
……
a - a = -9*/2^n
以上各等式相加
a - a2 = -9*(1/2^3 + 1/2^5 + …… + 1/2^n)
a + 5/2 = -9 * (1/2^3) * {1 - 1/4^[(n-1)/2)]/(1 - 1/4) = (-3/2)*[1 - 1/2^(n-1)]
a = -4 + 3/2^n
其中n為奇數(shù)
綜上所述,
對(duì)于數(shù)列奇數(shù)位
an = 4 - 3/2^(n-1)
對(duì)于數(shù)列偶數(shù)位
an = -4 + 3/2^(n-1)
二者可以合并寫(xiě)為
an = (-1)^n * [3/2^(n-1) -4]

我來(lái)回答

類似推薦

猜你喜歡

精品偷拍一区二区三区,亚洲精品永久 码,亚洲综合日韩精品欧美国产,亚洲国产日韩a在线亚洲

已知數(shù)列an的前n項(xiàng)和Sn滿足Sn-Sn-2=3(-1/2)^(n-1)(n>=3),且S1=1,S2=-3/2,求數(shù)列an

精品偷拍一区二区三区,亚洲精品永久码,亚洲综合日韩精品欧美国产,亚洲国产日韩a在线亚洲