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\(B=\dfrac{1}{2}+\dfrac{1}{6}+\dfrac{1}{12}+...+\dfrac{1}{9900}\)
\(B=\dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+...+\dfrac{1}{99.100}\)
\(B=\dfrac{1}{1}-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{99}-\dfrac{1}{100}\)
\(B=1-\dfrac{1}{100}\)
\(B=\dfrac{99}{100}\)
Vậy \(B=\dfrac{99}{100}\)
B \(=\) \(\dfrac{1}{2}\) \(+\) \(\dfrac{1}{6}\) \(+\) \(\dfrac{1}{12}\) \(+\) \(\dfrac{1}{20}\) \(+\) \(\dfrac{1}{30}\) \(+\) . . . . . \(+\) \(\dfrac{1}{9900}\)
\(=\) \(\dfrac{1}{1.2}\) \(+\) \(\dfrac{1}{2.3}\) \(+\) \(\dfrac{1}{3.4}\) \(+\) \(\dfrac{1}{4.5}\) \(+\) \(\dfrac{1}{5.6}\) \(+\) . . . . . \(+\) \(\dfrac{1}{99.100}\)
\(=\) \(\dfrac{1}{1}\) \(-\) \(\dfrac{1}{2}\) \(+\) \(\dfrac{1}{2}\) \(-\) \(\dfrac{1}{3}\) \(+\) \(\dfrac{1}{3}\) \(-\) \(\dfrac{1}{4}\) \(+\) \(\dfrac{1}{4}\) \(-\) \(\dfrac{1}{5}\) \(+\) \(\dfrac{1}{5}\) \(-\) \(\dfrac{1}{6}\) \(+\) . . . . . \(+\) \(\dfrac{1}{99}\) \(-\) \(\dfrac{1}{100}\)
\(=\) \(\dfrac{1}{1}\) \(-\) \(\dfrac{1}{100}\)
\(=\) \(\dfrac{99}{100}\)
ab-ac+bc-c2=b(a+c)-c(a+c)=(b-c)(a+c)
=>\(\orbr{\begin{cases}b=c+1,a=-1-c\\b=c-1,a=1-c\end{cases}}\)
\(\Leftrightarrow\frac{a}{b}=-1\)
A={5k+1} đk k thuộc N
học tốt
......................................
\(N=1+2+2^2+...+2^{2008}\)
\(\Leftrightarrow2N=2+2^2+...+2^{2009}\)
\(\Leftrightarrow N=2^{2009}-1\)
\(M=\dfrac{2^{2009}-1}{1-2^{2009}}=-1\)
A=-(1+2+3+......+2010)
A=-2011.2010:2=-(1005.2011)