tim a: 4/3*7+4/7*11+4/11*15+...+4/a*(a+4)=664/1995
b a+a/3+a/6+a/10+...+a/45\165/178
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\(S=\frac{4}{3\times7}+\frac{4}{7\times11}+\frac{4}{11\times15}+...+\frac{4}{\left(4x-1\right)\times\left(4x+3\right)}\)
\(=\frac{7-3}{3\times7}+\frac{11-4}{7\times11}+\frac{15-11}{11\times15}+...+\frac{\left(4x+3\right)-\left(4x-1\right)}{\left(4x-1\right)\times\left(4x+3\right)}\)
\(=\frac{1}{3}-\frac{1}{7}+\frac{1}{7}-\frac{1}{11}+\frac{1}{11}-\frac{1}{15}+...+\frac{1}{4x-1}-\frac{1}{4x+3}\)
\(=\frac{1}{3}-\frac{1}{4x+3}=\frac{664}{1995}\)
\(\Leftrightarrow\frac{1}{4x+3}=\frac{1}{1995}\)
\(\Leftrightarrow4x+3=1995\)
\(\Leftrightarrow x=498\).
Số hạng cuối cùng của dãy \(S\)là: \(\frac{1}{1991\times1995}\).
Tổng \(S\)có \(498\)số hạng.
a) \(\frac{7}{2}-\frac{14}{3}+\left(\frac{3}{4}-\frac{7}{3}\right)-\left(\frac{5}{6}-\frac{7}{4}\right)+\frac{11}{2}-3\)
\(=\frac{7}{2}-\frac{14}{3}+\frac{3}{4}-\frac{7}{3}-\frac{5}{6}+\frac{7}{4}+\frac{11}{2}-3\)
\(=\left(\frac{7}{2}+\frac{11}{2}\right)-\left(\frac{14}{3}+\frac{7}{3}\right)-3+\left(\frac{3}{4}+\frac{7}{4}\right)-\frac{5}{6}\)
\(=9-7-3+(\frac{5}{2}-\frac{5}{6})\)
\(=-1+\frac{5}{3}\)
\(=\frac{2}{3}\)
b) \(\frac{7}{3}-\frac{7}{5}+\frac{11}{10}-\left(\frac{2}{5}-\frac{5}{6}\right)+\frac{4}{15}-\frac{4}{3}\)
\(=\left(\frac{7}{3}-\frac{4}{3}\right)-\left(\frac{7}{5}+\frac{2}{5}\right)+(\frac{11}{10}+\frac{5}{6}+\frac{4}{15})\)
\(=1-\frac{9}{5}+\frac{11}{5}\)
\(=1-\left(\frac{9}{5}-\frac{11}{5}\right)\)
\(=1-\left(\frac{-2}{5}\right)\)
\(=1\frac{2}{5}\)
......................?
mik ko biết
mong bn thông cảm
nha ................
\(\frac{4}{3.7}+\frac{4}{7.11}+\frac{4}{11.15}+...+\frac{4}{a\left(a+4\right)}\)
\(=\frac{1}{3}-\frac{1}{7}+\frac{1}{7}-\frac{1}{11}+\frac{1}{11}-\frac{1}{15}+...+\frac{1}{a}-\frac{1}{a+4}\)
\(=\frac{1}{3}-\frac{1}{a+4}\)