Tính: $\frac{2.2014}{1+\frac{1}{1+2}+\frac{1}{1+2+3}+\frac{1}{1+2+3+4}+…+\frac{1}{1+2+3+2014}}$ 16/09/2021 Bởi Peyton Tính: $\frac{2.2014}{1+\frac{1}{1+2}+\frac{1}{1+2+3}+\frac{1}{1+2+3+4}+…+\frac{1}{1+2+3+2014}}$
Ta có $1 + 2 + \cdots + n = \dfrac{n(n+1)}{2}$ Do đó $\dfrac{2.2014}{1 + \frac{1}{1+2} + \frac{1}{1+ 2 + 3} + \cdots + \frac{1}{1 + 2 + \cdots + 2014}} = \dfrac{2.2014}{1 + \frac{2}{2.3} + \frac{2}{3.4} + \cdots + \frac{2}{2014.2015}}$ $= \dfrac{2.2014}{1 + 2 \left( \frac{1}{2.3} + \frac{1}{3.4} + \cdots + \frac{1}{2014.2015} \right)}$ $= \dfrac{2.2014}{1 + 2 \left( \frac{1}{2} – \frac{1}{3} + \frac{1}{3} – \frac{1}{4} +\cdots + \frac{1}{2014} – \frac{1}{2015} \right)}$ $= \dfrac{2.2014}{1 + 2 \left( \frac{1}{2} – \frac{1}{2015} \right)}$ $= \dfrac{2.2014}{1 + 2 . \frac{2013}{2.2015}}$ $= \dfrac{2.2014}{1 + \frac{2013}{2015}}$ $= \dfrac{4028}{\frac{4028}{2015}} = 2015$ Vậy biểu thức trên bằng $2015$. Bình luận
Ta có
$1 + 2 + \cdots + n = \dfrac{n(n+1)}{2}$
Do đó
$\dfrac{2.2014}{1 + \frac{1}{1+2} + \frac{1}{1+ 2 + 3} + \cdots + \frac{1}{1 + 2 + \cdots + 2014}} = \dfrac{2.2014}{1 + \frac{2}{2.3} + \frac{2}{3.4} + \cdots + \frac{2}{2014.2015}}$
$= \dfrac{2.2014}{1 + 2 \left( \frac{1}{2.3} + \frac{1}{3.4} + \cdots + \frac{1}{2014.2015} \right)}$
$= \dfrac{2.2014}{1 + 2 \left( \frac{1}{2} – \frac{1}{3} + \frac{1}{3} – \frac{1}{4} +\cdots + \frac{1}{2014} – \frac{1}{2015} \right)}$
$= \dfrac{2.2014}{1 + 2 \left( \frac{1}{2} – \frac{1}{2015} \right)}$
$= \dfrac{2.2014}{1 + 2 . \frac{2013}{2.2015}}$
$= \dfrac{2.2014}{1 + \frac{2013}{2015}}$
$= \dfrac{4028}{\frac{4028}{2015}} = 2015$
Vậy biểu thức trên bằng $2015$.