GIÚP MK VS
Tính Tổng:
a) $\frac{1}{1X2}$+ $\frac{1}{2X3}$ +…+$\frac{1}{101X102}$
b) $\frac{2}{1X3}$+ $\frac{2}{3X5}$+…+ $\frac{2}{49X51}$
c) $\frac{3}{3X8}$ + $\frac{3}{8X13}$+…+$\frac{3}{103X108}$
d)$\frac{2}{4X7}$+ $\frac{2}{7X10}$+…+ $\frac{2}{124X127}$
Đáp án:
\(\begin{array}{l}
a.\frac{{101}}{{102}}\\
b.\frac{{50}}{{51}}\\
c.\frac{7}{{36}}\\
d.\frac{{41}}{{254}}
\end{array}\)
Giải thích các bước giải:
\(\begin{array}{l}
a.\frac{1}{{1.2}} + \frac{1}{{2.3}} + … + \frac{1}{{101.102}}\\
= 1 – \frac{1}{2} + \frac{1}{2} – \frac{1}{3} + … + \frac{1}{{101}} – \frac{1}{{102}}\\
= 1 – \frac{1}{{102}} = \frac{{101}}{{102}}\\
b.\frac{2}{{1.3}} + \frac{2}{{3.5}} + … + \frac{2}{{49.51}}\\
= 1 – \frac{1}{3} + \frac{1}{3} – \frac{1}{5} + … + \frac{1}{{49}} – \frac{1}{{51}}\\
= 1 – \frac{1}{{51}} = \frac{{50}}{{51}}\\
c.\frac{3}{{3.8}} + \frac{3}{{8.13}} + … + \frac{3}{{103.108}}\\
= 3.\frac{1}{5}.\left( {\frac{5}{{3.8}} + \frac{5}{{8.13}} + … + \frac{5}{{103.108}}} \right)\\
= \frac{3}{5}\left( {\frac{1}{3} – \frac{1}{8} + \frac{1}{8} – \frac{1}{{13}} + … + \frac{1}{{103}} – \frac{1}{{108}}} \right)\\
= \frac{3}{5}.\left( {\frac{1}{3} – \frac{1}{{108}}} \right) = \frac{7}{{36}}\\
d.\frac{2}{{4.7}} + \frac{2}{{7.10}} + … + \frac{2}{{124.127}}\\
= 2.\frac{1}{3}.\left( {\frac{3}{{4.7}} + \frac{3}{{7.10}} + … + \frac{3}{{124.127}}} \right)\\
= \frac{2}{3}.\left( {\frac{1}{4} – \frac{1}{7} + \frac{1}{7} – \frac{1}{{10}} + … + \frac{1}{{124}} – \frac{1}{{127}}} \right)\\
= \frac{2}{3}.\left( {\frac{1}{4} – \frac{1}{{127}}} \right) = \frac{{41}}{{254}}
\end{array}\)