[toán 7] Tính $ m-n?$

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Tính $ m-n?$
$m=\frac{3}{2}+\frac{13}{12}+\frac{31}{30}+...+ \frac{9901}{9900}$
$n=50-(\frac{1}{2}.3+\frac{1}{4}.5+\frac{1}{6}.7+...+ \frac{1}{100}.101)$

 

[thieukhang61]

\[\begin{array}{l}
N = 50 - (\frac{1}{2}.3 + \frac{1}{4}.5 + \frac{1}{6}.7 + ... + \frac{1}{{100}}.101)\\
= 50 - (\frac{3}{2} + \frac{5}{4} + \frac{7}{6} + ... + \frac{{101}}{{100}})\\
= 1 - \frac{3}{2} + 1 - \frac{5}{4} + 1 - \frac{7}{6} + ... + 1 - \frac{{101}}{{100}}\\
= \frac{{ - 1}}{2} + \frac{{ - 1}}{4} + \frac{{ - 1}}{6} + ... + \frac{{ - 1}}{{100}}\\
M = \frac{3}{2} + \frac{{13}}{{12}} + \frac{{31}}{{30}} + ... + \frac{{9901}}{{9900}}\\
= \frac{{1.2 + 1}}{{1.2}} + \frac{{3.4 + 1}}{{3.4}} + \frac{{5.6 + 1}}{{5.6}} + ... + \frac{{99.100 + 1}}{{99.100}}\\
M - N = (\frac{{1.2 + 1}}{{1.2}} + \frac{{3.4 + 1}}{{3.4}} + \frac{{5.6 + 1}}{{5.6}} + ... + \frac{{99.100 + 1}}{{99.100}}) - (\frac{{ - 1}}{2} + \frac{{ - 1}}{4} + \frac{{ - 1}}{6} + ... + \frac{{ - 1}}{{100}})\\
= \frac{{1.2 + 1}}{{1.2}} + \frac{{3.4 + 1}}{{3.4}} + \frac{{5.6 + 1}}{{5.6}} + ... + \frac{{99.100 + 1}}{{99.100}} + \frac{1}{2} + \frac{1}{4} + \frac{1}{6} + ... + \frac{1}{{100}}\\
= \frac{{1.2 + 1}}{{1.2}} + \frac{1}{2} + \frac{{3.4 + 1}}{{3.4}} + \frac{1}{4} + \frac{{5.6 + 1}}{{5.6}} + \frac{1}{6} + ... + \frac{{99.100 + 1}}{{99.100}} + \frac{1}{{100}}\\
Lai\,\,co:\,\,\frac{{n(n + 1) + 1}}{{n(n + 1)}} + \frac{1}{{n + 1}} = \frac{{n(n + 1) + 1}}{{n(n + 1)}} + \frac{n}{{n(n + 1)}} = \frac{{n(n + 1) + 1 + n}}{{n(n + 1)}} = \frac{{{n^2} + 2n + 1}}{{n(n + 1)}} = \frac{{{{(n + 1)}^2}}}{{n(n + 1)}} = \frac{{n + 1}}{n}\\
= > M - N = 2 + \frac{4}{3} + \frac{6}{5} + ... + \frac{{100}}{{99}}\\
= 51 + \frac{1}{3} + \frac{1}{5} + ... + \frac{1}{{99}}\\
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\end{array}\]

 

[thieukhang61]

\[\begin{array}{l}
Cach\,\,khac:\\
M = \frac{3}{2} + \frac{{13}}{{12}} + \frac{{31}}{{30}} + ... + \frac{{9901}}{{9900}}\\
= 50 + \frac{1}{2} + \frac{1}{{12}} + \frac{1}{{30}} + ... + \frac{1}{{9900}}\\
N = 50 - (\frac{1}{2}.3 + \frac{1}{4}.5 + \frac{1}{6}.7 + ... + \frac{1}{{100}}.101)\\
= 50 - (\frac{3}{2} + \frac{5}{4} + \frac{7}{6} + ... + \frac{{101}}{{100}})\\
= 50 - \frac{3}{2} - \frac{5}{4} - \frac{7}{6} - ... - \frac{{101}}{{100}}\\
M - N = (50 + \frac{1}{2} + \frac{1}{{12}} + \frac{1}{{30}} + ... + \frac{1}{{9900}}) - (50 - \frac{3}{2} - \frac{5}{4} - \frac{7}{6} - ... - \frac{{101}}{{100}})\\
{\rm{ = }}50 + \frac{1}{2} + \frac{1}{{12}} + \frac{1}{{30}} + ... + \frac{1}{{9900}} - 50 + \frac{3}{2} + \frac{5}{4} + \frac{7}{6} + ... + \frac{{101}}{{100}}\\
= \frac{1}{2} + \frac{1}{{12}} + \frac{1}{{30}} + ... + \frac{1}{{9900}} + \frac{3}{2} + \frac{5}{4} + \frac{7}{6} + ... + \frac{{101}}{{100}}\\
= \frac{1}{2} + \frac{3}{2} + \frac{1}{{12}} + \frac{5}{4} + \frac{1}{{30}} + \frac{7}{6} + ... + \frac{1}{{9900}} + \frac{{101}}{{100}}\\
= \frac{1}{{1.2}} + \frac{3}{2} + \frac{1}{{3.4}} + \frac{5}{4} + \frac{1}{{5.6}} + \frac{7}{6} + ... + \frac{1}{{99.100}} + \frac{{101}}{{100}}\\
Lai\,\,co:\,\,\frac{1}{{n(n + 1)}} + \frac{{n + 2}}{{n + 1}} = \frac{1}{{n(n + 1)}} + \frac{{n(n + 2)}}{{n(n + 1)}} = \frac{{n(n + 2) + 1}}{{n(n + 1)}} = \frac{{{n^2} + 2n + 1}}{{n(n + 1)}} = \frac{{{{(n + 1)}^2}}}{{n(n + 1)}} = \frac{{n + 1}}{n}\\
= > M - N = 2 + \frac{4}{3} + \frac{6}{5} + ... + \frac{{100}}{{99}}\\
= 51 + \frac{1}{3} + \frac{1}{5} + ... + \frac{1}{{99}}\\

\end{array}\]