Tính tổng S = C2018^0 + 1/2C2018^1 + 1/3C2018^2 + ... + 1//2018C2018^2017 + 1/2019C2018^2018 A. S = 2^2018 + 1/2019 B. S = 2^2018 - 1/2019 + 1 C. S = 2^2019 - 1/2019 D. S = 2^
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24/04/2024
Tính tổng \[S = C_{2018}^0 + \frac{1}{2}C_{2018}^1 + \frac{1}{3}C_{2018}^2 + ... + \frac{1}{{2018}}C_{2018}^{2017} + \frac{1}{{2019}}C_{2018}^{2018}\]
A. \[S = \frac{{{2^{2018}} + 1}}{{2019}}\]
B. \[S = \frac{{{2^{2018}} - 1}}{{2019}} + 1\]
C. \[S = \frac{{{2^{2019}} - 1}}{{2019}}\]
D. \[S = \frac{{{2^{2018}} - 1}}{{2019}} - 1\]
Trả lời
Đáp án C
Phương pháp:
Tính tổng 2019S bằng cách nhận xét số hạng tổng quát của tổng này
Cách giải:
Ta có: \[S = C_{2018}^0 + \frac{1}{2}C_{2018}^1 + \frac{1}{3}C_{2018}^2 + ... + \frac{1}{{2018}}C_{2018}^{2017} + \frac{1}{{2019}}C_{2018}^{2018} = \sum\limits_{K = 0}^{2018} {\frac{1}{{k + 1}}C_{2018}^k} \]
\[ \Rightarrow 2019S = \sum\limits_{k = 0}^{2018} {\frac{{2019}}{{k + 1}}} C_{2018}^k = \sum\limits_{k = 0}^{2018} {\frac{{2019}}{{k + 1}}} .\frac{{2018!}}{{k!\left( {2018 - k} \right)!}} = \sum\limits_{k = 0}^{2018} {\frac{{2019!}}{{\left( {k + 1} \right)!\left( {2019 - \left( {k + 1} \right)} \right)!}}} = \sum\limits_{k = 0}^{2018} {C_{2019}^{k + 1}} \]
\[ = C_{2019}^1 + C_{2019}^2 + ... + C_{2019}^{2019} \Rightarrow 2019S + C_{2019}^0 + C_{2019}^1 + C_{2019}^2 + ... + C_{2019}^{2019} = {2^{2019}}\]
\[ \Rightarrow 2019 = {2^{2019}} - C_{2019}^0 = {2^{2019}} - 1 \Rightarrow S = \frac{{{2^{2019}} - 1}}{{2019}}\]
