ให้ $a_1,a_2,a_3,\cdots,a_{20}$ เป็นดับเลขคณิต และมีผลต่างร่วมเท่ากับ $\dfrac{2}{21}$ แล้ว

$$\frac{1}{21\left(a_{20}-a_1\right)}+\frac{1}{19\left(a_{19}-a_2\right)}+\frac{1}{17\left(a_{18}-a_3\right)}+\cdots+\frac{1}{5\left(a_{12}-a_9\right)}+\frac{1}{3\left(a_{11}-a_{10}\right)}$$

มีค่าเท่ากับเท่าใดต่อไปนี้

เฉลยละเอียด

[STEP]พิจารณาผลต่างร่วม[/STEP]

$\displaystyle a_{20} - a_1 = (a_1 + 19d) - a_1 = 19d$

$\displaystyle a_{19} - a_2 = (a_1 + 18d) - (a_1 + d) = 17d$

$\displaystyle a_{18} - a_3 = (a_1 + 17d) - (a_1 + 2d) = 15d$

                 $\vdots$

$\displaystyle a_{11} - a_{10} = (a_1 + 10d) - (a_1 + 9d) = d$

[STEP]หาคำตอบ[/STEP]

\begin{eqnarray*}
& & \frac{1}{21\left(a_{20}-a_1\right)}+\frac{1}{19\left(a_{19}-a_2\right)}+\frac{1}{17\left(a_{18}-a_3\right)}+\cdots+\frac{1}{5\left(a_{12}-a_9\right)}+\frac{1}{3\left(a_{11}-a_{10}\right)}\\
&=& \frac{1}{21\left(19d\right)}+\frac{1}{19\left(17d\right)}+\frac{1}{17\left(15d\right)}+\cdots+\frac{1}{5\left(3d\right)}+\frac{1}{3\left(d\right)}\\
&=& \frac{1}{d} \left ( \frac{1}{21 \times 19}+\frac{1}{19 \times 17}+\frac{1}{17 \times 15}+\cdots+\frac{1}{5 \times 3}+\frac{1}{3 \times 1} \right )
\end{eqnarray*}

มองเป็นอนุกรมเทเลสโคปิก จะได้

\begin{eqnarray*}
& & \frac{1}{d} \left ( \frac{1}{21 \times 19}+\frac{1}{19 \times 17}+\frac{1}{17 \times 15}+\cdots+\frac{1}{5 \times 3}+\frac{1}{3 \times 1} \right )\\
&=& \frac{2}{2} \times \frac{1}{d} \left ( \frac{1}{21 \times 19}+\frac{1}{19 \times 17}+\frac{1}{17 \times 15}+\cdots+\frac{1}{5 \times 3}+\frac{1}{3 \times 1} \right )\\
&=& \frac{1}{2d} \left ( \frac{2}{21 \times 19}+\frac{2}{19 \times 17}+\frac{2}{17 \times 15}+\cdots+\frac{2}{5 \times 3}+\frac{2}{3 \times 1} \right )
\end{eqnarray*}

ซึ่งสามารถแยกได้เป็น

\begin{eqnarray*}
& &\frac{1}{2d} \left ( \frac{1}{19} - \frac{1}{21} +\frac{1}{17} - \frac{1}{19} + \frac{1}{15} - \frac{1}{17} +\cdots+\frac{1}{3} - \frac{1}{5}+\frac{1}{1} - \frac{1}{3} \right )\\
&=& \frac{1}{2d} \left ( \cancel{\frac{1}{19}} - \frac{1}{21} +\cancel{\frac{1}{17}} - \cancel{\frac{1}{19}} + \cancel{\frac{1}{15}} - \cancel{\frac{1}{17}} +\cdots+\cancel{\frac{1}{3}} - \cancel{\frac{1}{5}}+\frac{1}{1} - \cancel{\frac{1}{3}} \right )\\
&=& \frac{1}{2d} \left( 1 - \frac{1}{21} \right)\\
&=& \frac{1}{2d} \left( \frac{20}{21} \right)
\end{eqnarray*}

แทนค่า $d = \dfrac{2}{21}$ จะได้

\begin{eqnarray*}
\frac{1}{2d} \left( \frac{20}{21} \right) &=& \frac{21}{4} \left( \frac{20}{21} \right)\\
&=& \frac{\cancel{21}}{\cancelto{1}{4}} \left( \frac{\cancelto{5}{20}}{\cancel{21}} \right)\\
&=& 5
\end{eqnarray*}

 

[ANS] $5$ [/ANS]