Cho \(A = \dfrac{1}{{{5^2}}} + \dfrac{2}{{{5^3}}} + \dfrac{3}{{{5^4}}} + \ldots + \dfrac{n}{{{5^{n + 1}}}} + \ldots + \dfrac{{11}}{{{5^{12}}}}\) với \(n \in \mathbb{N}\). So sánh \(A\) và \(\dfrac{1}{{16}} \cdot \)
Câu 386911: Cho \(A = \dfrac{1}{{{5^2}}} + \dfrac{2}{{{5^3}}} + \dfrac{3}{{{5^4}}} + \ldots + \dfrac{n}{{{5^{n + 1}}}} + \ldots + \dfrac{{11}}{{{5^{12}}}}\) với \(n \in \mathbb{N}\). So sánh \(A\) và \(\dfrac{1}{{16}} \cdot \)
A. \(A > \dfrac{1}{{16}}\)
B. \(A < \dfrac{1}{{16}}\)
C. \(A = \dfrac{1}{{16}}\)
D. Không có đáp án
Tính tổng \(A\)
Bước 1: Mẫu là lũy thừa cơ số \(5\) nên để tính tổng \(A\) ta nhân cả hai vế với \(5\).
Bước 2: Lấy \(5A - A\) Suy ra tổng \(A\).
Bước 3: Viết \(A = \dfrac{1}{{16}} \cdot a\) với \(a \ne 0\). Nếu \(a < 1\) thì \(A < \dfrac{1}{{16}}\) và ngược lại.
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Đáp án : B(0) bình luận (0) lời giải
Giải chi tiết:
\(A = \dfrac{1}{{{5^2}}} + \dfrac{2}{{{5^3}}} + \dfrac{3}{{{5^4}}} + \ldots + \dfrac{n}{{{5^{n + 1}}}} + \ldots + \dfrac{{11}}{{{5^{12}}}}\)
\( \Rightarrow 5A = 5 \cdot \left( {\dfrac{1}{{{5^2}}} + \dfrac{2}{{{5^3}}} + \dfrac{3}{{{5^4}}} + \ldots + \dfrac{n}{{{5^{n + 1}}}} + \ldots + \dfrac{{11}}{{{5^{12}}}}} \right)\)\( = \dfrac{1}{5} + \dfrac{2}{{{5^2}}} + \dfrac{3}{{{5^3}}} + \ldots + \dfrac{n}{{{5^n}}} + \ldots + \dfrac{{11}}{{{5^{11}}}}\)
\(\begin{array}{l} \Rightarrow 5A = \dfrac{1}{5} + \dfrac{2}{{{5^2}}} + \dfrac{3}{{{5^3}}} + \ldots + \dfrac{n}{{{5^n}}} + \ldots + \dfrac{{11}}{{{5^{11}}}}\\\,\,\,\,\,\,\,\,A = \dfrac{1}{{{5^2}}} + \dfrac{2}{{{5^3}}} + \dfrac{3}{{{5^4}}} + \ldots + \dfrac{n}{{{5^{n + 1}}}} + \ldots + \dfrac{{11}}{{{5^{12}}}}\\ \Rightarrow 5A - A = \left( {\dfrac{1}{5} + \dfrac{2}{{{5^2}}} + \dfrac{3}{{{5^3}}} + \ldots + \dfrac{n}{{{5^n}}} + \ldots + \dfrac{{11}}{{{5^{11}}}}} \right) - \left( {\dfrac{1}{{{5^2}}} + \dfrac{2}{{{5^3}}} + \dfrac{3}{{{5^4}}} + \ldots + \dfrac{n}{{{5^{n + 1}}}} + \ldots + \dfrac{{11}}{{{5^{12}}}}} \right)\\\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = \dfrac{1}{5} + \dfrac{2}{{{5^2}}} + \dfrac{3}{{{5^3}}} + \ldots + \dfrac{n}{{{5^n}}} + \ldots + \dfrac{{11}}{{{5^{11}}}} - \dfrac{1}{{{5^2}}} - \dfrac{2}{{{5^3}}} - \dfrac{3}{{{5^4}}} - \ldots - \dfrac{n}{{{5^{n + 1}}}} - \ldots - \dfrac{{11}}{{{5^{12}}}}\\\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = \dfrac{1}{5} + \left( {\dfrac{2}{{{5^2}}} - \dfrac{1}{{{5^2}}}} \right) + \left( {\dfrac{3}{{{5^3}}} - \dfrac{2}{{{5^3}}}} \right) + \ldots + \left( {\dfrac{n}{{{5^n}}} - \dfrac{{n - 1}}{{{5^n}}}} \right) + \ldots + \left( {\dfrac{{11}}{{{5^{11}}}} - \dfrac{{10}}{{{5^{11}}}}} \right) - \dfrac{{11}}{{{5^{12}}}}\\\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = \dfrac{1}{5} + \dfrac{1}{{{5^2}}} + \dfrac{1}{{{5^3}}} + \ldots + \dfrac{1}{{{5^n}}} + \ldots + \dfrac{1}{{{5^{11}}}} - \dfrac{{11}}{{{5^{12}}}}\end{array}\)
\( \Rightarrow 4A = \dfrac{1}{5} + \dfrac{1}{{{5^2}}} + \dfrac{1}{{{5^3}}} + \ldots + \dfrac{1}{{{5^n}}} + \ldots + \dfrac{1}{{{5^{11}}}} - \dfrac{{11}}{{{5^{12}}}}\)
\( \Rightarrow 4A = \left( {\dfrac{1}{5} + \dfrac{1}{{{5^2}}} + \dfrac{1}{{{5^3}}} + \ldots + \dfrac{1}{{{5^n}}} + \ldots + \dfrac{1}{{{5^{11}}}}} \right) - \dfrac{{11}}{{{5^{12}}}}\)
Đặt \(B = \dfrac{1}{5} + \dfrac{1}{{{5^2}}} + \dfrac{1}{{{5^3}}} + \ldots + \dfrac{1}{{{5^n}}} + \ldots + \dfrac{1}{{{5^{11}}}}\).
Biểu thức \(A\) trở thành: \(4A = B - \dfrac{{11}}{{{5^{12}}}}\)
Tính tổng: \(B = \dfrac{1}{5} + \dfrac{1}{{{5^2}}} + \dfrac{1}{{{5^3}}} + \ldots + \dfrac{1}{{{5^n}}} + \ldots + \dfrac{1}{{{5^{11}}}}\)
\( \Rightarrow 5B = 5 \cdot \left( {\dfrac{1}{5} + \dfrac{1}{{{5^2}}} + \dfrac{1}{{{5^3}}} + \ldots + \dfrac{1}{{{5^n}}} + \ldots + \dfrac{1}{{{5^{11}}}}} \right)\)
\( \Rightarrow 5B = 1 + \dfrac{1}{5} + \dfrac{1}{{{5^2}}} + \dfrac{1}{{{5^3}}} + \ldots + \dfrac{1}{{{5^n}}} + \ldots + \dfrac{1}{{{5^{10}}}}\)
\(5B - B = \left( {1 + \dfrac{1}{5} + \dfrac{1}{{{5^2}}} + \dfrac{1}{{{5^3}}} + \ldots + \dfrac{1}{{{5^n}}} + \ldots + \dfrac{1}{{{5^{10}}}}} \right) - \left( {\dfrac{1}{5} + \dfrac{1}{{{5^2}}} + \dfrac{1}{{{5^3}}} + \ldots + \dfrac{1}{{{5^n}}} + \ldots + \dfrac{1}{{{5^{11}}}}} \right)\)
\(\begin{array}{l} \Rightarrow 4B = 1 + \dfrac{1}{5} + \dfrac{1}{{{5^2}}} + \dfrac{1}{{{5^3}}} + \ldots + \dfrac{1}{{{5^n}}} + \ldots + \dfrac{1}{{{5^{10}}}} - \dfrac{1}{5} - \dfrac{1}{{{5^2}}} - \dfrac{1}{{{5^3}}} - \ldots - \dfrac{1}{{{5^n}}} - \ldots - \dfrac{1}{{{5^{11}}}}\\\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = 1 - \dfrac{1}{{{5^{11}}}} = \dfrac{{{5^{11}} - 1}}{{{5^{11}}}}\end{array}\)
\( \Rightarrow B = \dfrac{{{5^{11}} - 1}}{{{{4.5}^{11}}}}\)
Thay \(B = \dfrac{{{5^{11}} - 1}}{{{{4.5}^{11}}}}\) vào biểu thức \(A\) ta được:
\(\begin{array}{l}4A = \dfrac{{{5^{11}} - 1}}{{{{4.5}^{11}}}} - \dfrac{{11}}{{{5^{12}}}} = \dfrac{{5\left( {{5^{11}} - 1} \right)}}{{{{4.5}^{12}}}} - \dfrac{{44}}{{{{4.5}^{12}}}}\\\,\,\,\,\,\,\,\, = \dfrac{{{5^{12}} - 5 - 44}}{{{{4.5}^{12}}}} = \dfrac{{{5^{12}} - 49}}{{{{4.5}^{12}}}}.\end{array}\)
\( \Rightarrow A = \dfrac{{{5^{12}} - 49}}{{{{16.5}^{12}}}} = \dfrac{1}{{16}} \cdot \dfrac{{{5^{12}} - 49}}{{{5^{12}}}}\)\( = \dfrac{1}{{16}} \cdot \left( {\dfrac{{{5^{12}}}}{{{5^{12}}}} - \dfrac{{49}}{{{5^{12}}}}} \right) = \dfrac{1}{{16}} \cdot \left( {1 - \dfrac{{49}}{{{5^{12}}}}} \right)\)
Vì \(1 - \dfrac{{49}}{{{5^{12}}}} < 1 \Rightarrow \dfrac{1}{{16}} \cdot \left( {1 - \dfrac{{49}}{{{5^{12}}}}} \right) < \dfrac{1}{{16}} \Rightarrow A < \dfrac{1}{{16}}\)
Vậy \(A < \dfrac{1}{{16}} \cdot \)
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