\(A=1-2+\frac{1}{3}+4-5+\frac{1}{6}+...+2014-2015+\frac{1}{2016}\)
\(=\left(-1\right)+\frac{1}{3}+\left(-1\right)+\frac{1}{6}+...+\left(-1\right)+\frac{1}{2016}\)
\(=\left[\left(-1\right)+\left(-1\right)+...+\left(-1\right)\right]+\left(\frac{1}{3}+\frac{1}{6}+...+\frac{1}{2016}\right)\)
\(=\left(-1\right)\cdot685+2\left(\frac{1}{6}+\frac{1}{12}+...+\frac{1}{4032}\right)=-685+2\cdot\left(\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+...+\frac{1}{63\cdot64}\right)\)
\(=-685+2\cdot\left(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{63}-\frac{1}{64}\right)=-685+2\cdot\left(\frac{1}{2}-\frac{1}{64}\right)\)
\(=-685+2\cdot\left(\frac{32}{64}-\frac{1}{64}\right)=-685+2\cdot\frac{31}{64}=-685+\frac{31}{32}=-\frac{21889}{32}\)
bài này bn ko làm được à
để mk xem có làm được ko?
