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2023新高考Ⅰ卷數(shù)學(xué)逐題解析(6)

2023-06-16 00:27 作者:CHN_ZCY  | 我要投稿

封面:和泉紗霧(《埃羅芒阿老師》)


20. 設(shè)等差數(shù)列%5Cleft%5C%7Ba_n%5Cright%5C%7D的公差為d,且d%3E1. 令b_n%3D%5Cfrac%7Bn%5E2%2Bn%7D%7Ba_n%7D,記S_n,T_n分別為數(shù)列%5Cleft%5C%7Ba_n%5Cright%5C%7D%5Cleft%5C%7Bb_n%5Cright%5C%7D的前n項和.

(1)若3a_2%3D3a_1%2Ba_3,S_3%2BT_3%3D21,求%5Cleft%5C%7Ba_n%5Cright%5C%7D的通項公式;

(2)若%5Cleft%5C%7Bb_n%5Cright%5C%7D為等差數(shù)列,且S_%7B99%7D-T_%7B99%7D%3D99,求d.

答案? (1)a_n%3D3n

(2)%5Cfrac%7B51%7D%7B50%7D.

解析??本題考察等差數(shù)列的定義、通項公式和求和公式,屬于中檔題.

(1)由3a_2%3D3a_1%2Ba_3,得3a_1%2B3d%3D3a_1%2Ba_1%2B2d,得a_1%3Dd.

所以a_n%3Ddn,b_n%3D%5Cfrac%7Bn%2B1%7D%7Bd%7D.

S_3%2BT_3%3D21,得6d%2B%5Cfrac%7B9%7D%7Bd%7D%3D21,得2d%5E2-7d%2B3%3D0?%5Cleft(d%3E1%5Cright).

解得?d%3D%5Cfrac%7B1%7D%7B2%7D%E6%88%963.?由于d%3E1,所以d%3D3.

所以a_n%3D3n.

(2)? 設(shè)a_n%3Ddn%2Bt,則b_n%3D%5Cfrac%7Bn%5E2%2Bn%7D%7Bdn%2Bt%7D.

%5Cleft%5C%7Bb_n%5Cright%5C%7D為等差數(shù)列,得b_%7Bn%2B1%7D-b_n%3D%5Cfrac%7B1%7D%7Bd%7D%2B%5Cfrac%7Bt-%5Cfrac%7Bt%5E2%7D%7Bd%7D%7D%7Bd%5E2%20n%5E2%2B%5Cleft(2dt%2Bd%5E2%5Cright)n%2Bdt%2Bt%5E2%7D為定值.

t-%5Cfrac%7Bt%5E2%7D%7Bd%7D%3D0,即?t%3D0%E6%88%96d.

t%3D0,則a_n%3Ddn,b_n%3D%5Cfrac%7Bn%2B1%7D%7Bd%7D.

所以

99%3DS_%7B99%7D-T_%7B99%7D%3D%5Cfrac%7B99%7D%7B2%7D%5Cleft(d%2B99d%5Cright)-%5Cfrac%7B99%7D%7B2%7D%5Cleft(%5Cfrac%7B2%7D%7Bd%7D%2B%5Cfrac%7B100%7D%7Bd%7D%5Cright)%3D99%5Cleft(50d-%5Cfrac%7B51%7D%7Bd%7D%5Cright)

50d%5E2-d-51%3D0%5Cleft(d%3E1%5Cright).

解得d%3D-1%E6%88%96%5Cfrac%7B51%7D%7B50%7D.?由于d%3E1,所以d%3D%5Cfrac%7B51%7D%7B50%7D.

t%3Dd,則a_n%3Ddn%2Bd,b_n%3D%5Cfrac%7Bn%7D%7Bd%7D.

所以

99%3DS_%7B99%7D-T_%7B99%7D%3D%5Cfrac%7B99%7D%7B2%7D%5Cleft(2d%2B100d%5Cright)-%5Cfrac%7B99%7D%7B2%7D%5Cleft(%5Cfrac%7B1%7D%7Bd%7D%2B%5Cfrac%7B99%7D%7Bd%7D%5Cright)%3D99%5Cleft(51d-%5Cfrac%7B50%7D%7Bd%7D%5Cright)

51d%5E2-d-50%3D0%5Cleft(d%3E1%5Cright).

解得d%3D1%E6%88%96-%5Cfrac%7B50%7D%7B51%7D.?由于d%3E1,所以這兩個解均舍去.

綜上,d%3D%5Cfrac%7B51%7D%7B50%7D.

21. 甲乙兩人投籃,每次由其中一人投籃,規(guī)則如下:若命中則此人繼續(xù)投籃,若未命中則換

為對方投籃. 無論之前投籃情況如何,甲每次投籃的命中率均為0.6,乙每次投籃的命中率均

0.8,由抽簽決定第一次投籃的人選,第一次投籃的人是甲,乙的概率各為0.5.

(1)求第2次投籃的人是乙的概率;

(2)求第i次投籃的人是甲的概率;

(3)已知:若隨機變量X_i服從兩點分布,且P%5Cleft(X_i%3D1%5Cright)%3D1-P%5Cleft(X_i%3D0%5Cright)%3Dq_i,i%3D1%2C2%2C%5Ccdots%2Cn,則E%5Cleft(%5Csum_%7Bi%3D1%7D%5En%20X_i%20%5Cright)%3D%5Csum_%7Bi%3D1%7D%5En%20q_i. 記前n次(即從第1次到第n次投籃)中甲投籃的次數(shù)為Y,求E%5Cleft(Y%5Cright).

答案? (1)%5Cfrac%7B3%7D%7B5%7D;

(2)%5Cfrac%7B1%7D%7B6%7D%20%5Ccdot%20%5Cleft(%5Cfrac%7B2%7D%7B5%7D%5Cright)%5E%7Bi-1%7D%2B%5Cfrac%7B1%7D%7B3%7D

(3)%5Cfrac%7B1%7D%7B3%7Dn%2B%5Cfrac%7B5%7D%7B18%7D-%5Cfrac%7B5%7D%7B18%7D%5Ccdot%5Cleft(%5Cfrac%7B2%7D%7B5%7D%5Cright)%5En.

解析? 本題考察事件與概率的運算,并要求借助數(shù)列解決概率問題,屬于中檔題.

(1)設(shè)A%3D%E2%80%9C%E7%AC%AC%E4%B8%80%E6%AC%A1%E6%8A%95%E7%AF%AE%E7%9A%84%E4%BA%BA%E6%98%AF%E7%94%B2%22,B%3D%E2%80%9C%E7%AC%AC%E4%B8%80%E6%AC%A1%E6%8A%95%E7%AF%AE%E7%9A%84%E4%BA%BA%E6%8A%95%E4%B8%AD%E2%80%9D,C%3D%E2%80%9C%E7%AC%AC%E4%BA%8C%E6%AC%A1%E6%8A%95%E7%AF%AE%E7%9A%84%E4%BA%BA%E6%98%AF%E4%B9%99%E2%80%9D.

P%5Cleft(C%5Cright)%3DP%5Cleft(A%20%5Cbar%7BB%7D%20%5Cright)%2BP%5Cleft(%20%5Cbar%7BA%7D%20B%20%5Cright)%3DP%5Cleft(A%5Cright)P%5Cleft(%5Cbar%7BB%7D%20%5Cvert%20A%20%5Cright)%2BP%5Cleft(%5Cbar%7BA%7D%5Cright)P%5Cleft(B%5Cvert%5Cbar%7BA%7D%5Cright)%3D%5Cfrac%7B1%7D%7B2%7D%5Ccdot%5Cleft(1-%5Cfrac%7B3%7D%7B5%7D%5Cright)%2B%5Cfrac%7B1%7D%7B2%7D%5Ccdot%5Cfrac%7B4%7D%7B5%7D%3D%5Cfrac%7B3%7D%7B5%7D.

(2)假設(shè)第i次投籃的人是甲的概率為p_i,則

p_i%3D%5Cfrac%7B3%7D%7B5%7Dp_%7Bi-1%7D%2B%5Cleft(1-%5Cfrac%7B4%7D%7B5%7D%5Cright)%5Cleft(1-p_%7Bi-1%7D%5Cright)%3D%5Cfrac%7B2%7D%7B5%7Dp_%7Bi-1%7D%2B%5Cfrac%7B1%7D%7B5%7D%5Cleft(i%5Cgeq2%5Cright).

p_i-%5Cfrac%7B1%7D%7B3%7D%3D%5Cfrac%7B2%7D%7B5%7D%5Cleft(p_%7Bi-1%7D-%5Cfrac%7B1%7D%7B3%7D%5Cright)%5Cleft(i%5Cgeq2%5Cright).

所以p_i-%5Cfrac%7B1%7D%7B3%7D%3D%5Cfrac%7B1%7D%7B6%7D%20%5Ccdot%20%5Cleft(%5Cfrac%7B2%7D%7B5%7D%5Cright)%5E%7Bi-1%7D.

所以p_i%3D%5Cfrac%7B1%7D%7B6%7D%20%5Ccdot%20%5Cleft(%5Cfrac%7B2%7D%7B5%7D%5Cright)%5E%7Bi-1%7D%2B%5Cfrac%7B1%7D%7B3%7D.

(3)設(shè)指示變量Y_i%3D%5Cleft%5C%7B%5Cbegin%7Baligned%7D%0A1%2C%E7%AC%ACi%E6%AC%A1%E6%8A%95%E7%AF%AE%E7%9A%84%E4%BA%BA%E6%98%AF%E7%94%B2%5C%5C%0A0%2C%E7%AC%ACi%E6%AC%A1%E6%8A%95%E7%AF%AE%E7%9A%84%E4%BA%BA%E4%B8%8D%E6%98%AF%E7%94%B2%0A%5Cend%7Baligned%7D%5Cright.,則Y_i服從兩點分布,且Y%3D%5Csum_%7Bi%3D1%7D%5En%20Y_i.

P%5Cleft(Y_i%3D1%5Cright)%3Dp_i.

所以

E%5Cleft(Y%5Cright)%3DE%5Cleft(%5Csum_%7Bi%3D1%7D%5En%20Y_i%5Cright)%3D%5Csum_%7Bi%3D1%7D%5En%20p_i%3D%5Csum_%7Bi%3D1%7D%5En%5Cleft%5B%5Cfrac%7B1%7D%7B6%7D%20%5Ccdot%20%5Cleft(%5Cfrac%7B2%7D%7B5%7D%5Cright)%5E%7Bi-1%7D%2B%5Cfrac%7B1%7D%7B3%7D%5Cright%5D%3D%5Cfrac%7B1%7D%7B3%7Dn%2B%5Cfrac%7B5%7D%7B18%7D-%5Cfrac%7B5%7D%7B18%7D%5Ccdot%5Cleft(%5Cfrac%7B2%7D%7B5%7D%5Cright)%5En.





2023新高考Ⅰ卷數(shù)學(xué)逐題解析(6)的評論 (共 條)

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