Toán Tính tổng sau : $\frac{1}{4}+\frac{2}{4^2}+\frac{3}{4^3}+…+\frac{2014}{4^{2014}}$ 19/07/2021 By Kaylee Tính tổng sau : $\frac{1}{4}+\frac{2}{4^2}+\frac{3}{4^3}+…+\frac{2014}{4^{2014}}$
Giải thích các bước giải: Ta có: \(\begin{array}{l}A = \frac{1}{4} + \frac{2}{{{4^2}}} + \frac{3}{{{4^3}}} + … + \frac{{2014}}{{{4^{2014}}}}\\ \Rightarrow 4A = 1 + \frac{2}{4} + \frac{3}{{{4^2}}} + \frac{4}{{{4^3}}} + …. + \frac{{2014}}{{{4^{2013}}}}\\ \Rightarrow 4A – A = \left( {1 + \frac{2}{4} + \frac{3}{{{4^2}}} + \frac{4}{{{4^3}}} + …. + \frac{{2014}}{{{4^{2013}}}}} \right) – \left( {\frac{1}{4} + \frac{2}{{{4^2}}} + \frac{3}{{{4^3}}} + … + \frac{{2014}}{{{4^{2014}}}}} \right)\\ \Leftrightarrow 3A = 1 + \frac{1}{4} + \frac{1}{{{4^2}}} + \frac{1}{{{4^3}}} + …. + \frac{1}{{{4^{2013}}}} – \frac{{2014}}{{{4^{2014}}}}\\B = 1 + \frac{1}{4} + \frac{1}{{{4^2}}} + \frac{1}{{{4^3}}} + …. + \frac{1}{{{4^{2013}}}}\\ \Rightarrow 4B = 4 + 1 + \frac{1}{4} + \frac{1}{{{4^2}}} + …. + \frac{1}{{{4^{2012}}}}\\ \Rightarrow 4B – B = \left( {4 + 1 + \frac{1}{4} + \frac{1}{{{4^2}}} + …. + \frac{1}{{{4^{2012}}}}} \right) – \left( {1 + \frac{1}{4} + \frac{1}{{{4^2}}} + \frac{1}{{{4^3}}} + …. + \frac{1}{{{4^{2013}}}}} \right)\\ \Leftrightarrow 3B = 4 – \frac{1}{{{4^{2013}}}} \Rightarrow B = \frac{4}{3} – \frac{1}{{{{3.4}^{2013}}}}\\3A = B – \frac{{2014}}{{{4^{2014}}}} = \frac{4}{3} – \left( {\frac{1}{{{{3.4}^{2013}}}} + \frac{{2014}}{{{4^{2014}}}}} \right) \Rightarrow A = ….\end{array}\) Trả lời
Giải thích các bước giải:
Ta có:
\(\begin{array}{l}
A = \frac{1}{4} + \frac{2}{{{4^2}}} + \frac{3}{{{4^3}}} + … + \frac{{2014}}{{{4^{2014}}}}\\
\Rightarrow 4A = 1 + \frac{2}{4} + \frac{3}{{{4^2}}} + \frac{4}{{{4^3}}} + …. + \frac{{2014}}{{{4^{2013}}}}\\
\Rightarrow 4A – A = \left( {1 + \frac{2}{4} + \frac{3}{{{4^2}}} + \frac{4}{{{4^3}}} + …. + \frac{{2014}}{{{4^{2013}}}}} \right) – \left( {\frac{1}{4} + \frac{2}{{{4^2}}} + \frac{3}{{{4^3}}} + … + \frac{{2014}}{{{4^{2014}}}}} \right)\\
\Leftrightarrow 3A = 1 + \frac{1}{4} + \frac{1}{{{4^2}}} + \frac{1}{{{4^3}}} + …. + \frac{1}{{{4^{2013}}}} – \frac{{2014}}{{{4^{2014}}}}\\
B = 1 + \frac{1}{4} + \frac{1}{{{4^2}}} + \frac{1}{{{4^3}}} + …. + \frac{1}{{{4^{2013}}}}\\
\Rightarrow 4B = 4 + 1 + \frac{1}{4} + \frac{1}{{{4^2}}} + …. + \frac{1}{{{4^{2012}}}}\\
\Rightarrow 4B – B = \left( {4 + 1 + \frac{1}{4} + \frac{1}{{{4^2}}} + …. + \frac{1}{{{4^{2012}}}}} \right) – \left( {1 + \frac{1}{4} + \frac{1}{{{4^2}}} + \frac{1}{{{4^3}}} + …. + \frac{1}{{{4^{2013}}}}} \right)\\
\Leftrightarrow 3B = 4 – \frac{1}{{{4^{2013}}}} \Rightarrow B = \frac{4}{3} – \frac{1}{{{{3.4}^{2013}}}}\\
3A = B – \frac{{2014}}{{{4^{2014}}}} = \frac{4}{3} – \left( {\frac{1}{{{{3.4}^{2013}}}} + \frac{{2014}}{{{4^{2014}}}}} \right) \Rightarrow A = ….
\end{array}\)