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Câu hỏi: Giá trị nào của tổng $4+44+444+...+44...4$ (tổng đó có 2018 số hạng) bằng
A. $\dfrac{4}{9}\left( \dfrac{{{10}^{2019}}-10}{9}-2018 \right)$.
B. $\dfrac{4}{9}\left( \dfrac{{{10}^{2019}}-10}{9}+2018 \right)$.
C. $\dfrac{4}{9}\left( {{10}^{2018}}-1 \right)$.
D. $\dfrac{40}{9}\left( {{10}^{2018}}-1 \right)+2018$.
A. $\dfrac{4}{9}\left( \dfrac{{{10}^{2019}}-10}{9}-2018 \right)$.
B. $\dfrac{4}{9}\left( \dfrac{{{10}^{2019}}-10}{9}+2018 \right)$.
C. $\dfrac{4}{9}\left( {{10}^{2018}}-1 \right)$.
D. $\dfrac{40}{9}\left( {{10}^{2018}}-1 \right)+2018$.
Xét dãy số có $\left\{ \begin{aligned}
& {{u}_{1}}=4 \\
& {{u}_{n+1}}=10{{u}_{n}}+4 \\
\end{aligned} \right.\Leftrightarrow \left\{ \begin{aligned}
& {{u}_{1}}=4 \\
& {{u}_{n+1}}+\dfrac{4}{9}=10\left( {{u}_{n}}+\dfrac{4}{9} \right) \\
\end{aligned} \right.$
Đặt ${{v}_{n}}={{u}_{n}}+\dfrac{4}{9}\Rightarrow \left\{ \begin{aligned}
& {{v}_{1}}=\dfrac{40}{9} \\
& {{v}_{n+1}}=10{{v}_{n}} \\
\end{aligned} \right.\Rightarrow v\left( n \right)$ là cấp số nhân.
Ta có: ${{S}_{n}}={{u}_{1}}+{{u}_{2}}+......+{{u}_{2018}}={{v}_{1}}-\dfrac{4}{9}+{{v}_{2}}-\dfrac{v}{9}...+{{v}_{2018}}-\dfrac{4}{9}={{v}_{1}}+{{v}_{2}}+...+{{v}_{2018}}-\dfrac{2018.4}{9}$
Trong đó ${{S}_{v\left( 2018 \right)}}=\dfrac{1-{{q}^{n}}}{1-q}.{{v}_{1}}=\dfrac{1-{{10}^{2018}}}{1-10}.\dfrac{40}{9}=\dfrac{40.\left( {{10}^{2018}}-1 \right)}{81}$
Vậy tổng là $S=\dfrac{40}{81}\left( {{10}^{2018}}-1 \right)-\dfrac{4}{9}.2018=\dfrac{4}{9}\left( \dfrac{{{10}^{2018}}-10}{9}-2018 \right)$.
& {{u}_{1}}=4 \\
& {{u}_{n+1}}=10{{u}_{n}}+4 \\
\end{aligned} \right.\Leftrightarrow \left\{ \begin{aligned}
& {{u}_{1}}=4 \\
& {{u}_{n+1}}+\dfrac{4}{9}=10\left( {{u}_{n}}+\dfrac{4}{9} \right) \\
\end{aligned} \right.$
Đặt ${{v}_{n}}={{u}_{n}}+\dfrac{4}{9}\Rightarrow \left\{ \begin{aligned}
& {{v}_{1}}=\dfrac{40}{9} \\
& {{v}_{n+1}}=10{{v}_{n}} \\
\end{aligned} \right.\Rightarrow v\left( n \right)$ là cấp số nhân.
Ta có: ${{S}_{n}}={{u}_{1}}+{{u}_{2}}+......+{{u}_{2018}}={{v}_{1}}-\dfrac{4}{9}+{{v}_{2}}-\dfrac{v}{9}...+{{v}_{2018}}-\dfrac{4}{9}={{v}_{1}}+{{v}_{2}}+...+{{v}_{2018}}-\dfrac{2018.4}{9}$
Trong đó ${{S}_{v\left( 2018 \right)}}=\dfrac{1-{{q}^{n}}}{1-q}.{{v}_{1}}=\dfrac{1-{{10}^{2018}}}{1-10}.\dfrac{40}{9}=\dfrac{40.\left( {{10}^{2018}}-1 \right)}{81}$
Vậy tổng là $S=\dfrac{40}{81}\left( {{10}^{2018}}-1 \right)-\dfrac{4}{9}.2018=\dfrac{4}{9}\left( \dfrac{{{10}^{2018}}-10}{9}-2018 \right)$.
Đáp án A.