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Đặt :
\(A=\dfrac{1}{2.5}+\dfrac{1}{5.8}+.........+\dfrac{1}{\left(3n-1\right)\left(3n+2\right)}\)
\(\Leftrightarrow3A=\dfrac{3}{2.5}+\dfrac{3}{5.8}+............+\dfrac{3}{\left(3n-1\right)\left(3n+2\right)}\)
\(\Leftrightarrow3A=\dfrac{1}{2}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{8}+........+\dfrac{1}{3n-1}-\dfrac{1}{3n+2}\)
\(\Leftrightarrow3A=\dfrac{1}{2}-\dfrac{1}{3n+2}\)
@Akai Haruma em không hiểu tại sao bài kia chị lại tick cho bạn đó ạ,đề nói chứng minh,mak bạn đó đã làm hết đâu:
\(VT=\dfrac{1}{2.5}+\dfrac{1}{5.8}+\dfrac{1}{8.11}+...+\dfrac{1}{\left(3n-1\right)\left(3n+2\right)}\)
\(VT=\dfrac{1}{3}\left(\dfrac{1}{2}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{8}+\dfrac{1}{8}-\dfrac{1}{11}+...+\dfrac{1}{3n-1}+\dfrac{1}{3n+2}\right)\)
\(VT=\dfrac{1}{3}\left(\dfrac{1}{2}-\dfrac{1}{3n+2}\right)\)
\(VT=\dfrac{1}{6}-\dfrac{1}{9n+6}\)
\(VT=\dfrac{9n+6}{54n+36}-\dfrac{6}{54n+36}\)
\(VT=\dfrac{9n+6-6}{54n+36}=\dfrac{9n}{54n+36}=\dfrac{9n}{9\left(6n+4\right)}=\dfrac{n}{6n+4}=VP\left(đpcm\right)\)
Ta có : A = 1/ 2.5 + 1/ 5.8 + 1/ 8.11 + ... + 1/ (3n-1).(3n+2) .
= 1/2 - 1/5 + 1/5 - 1/8 + 1/8 - 1/11 + ... + 1/ 3n-1 - 1/ 3n+2 .
= 1/2 - 1/ 3n+2 .
= 3n + 2 - 2 / 2 .( 3n+2 ) .
= 3n / 2.(3n+2) .
Giải:
\(C=\left(1-\dfrac{2}{2.3}\right)\left(1-\dfrac{2}{3.4}\right)\left(1-\dfrac{2}{4.5}\right)...\left(1-\dfrac{2}{n\left(n+1\right)}\right)\)
Đk: \(n\ne0;n\ne-1\)
\(C=\left(1-\dfrac{2}{2.3}\right)\left(1-\dfrac{2}{3.4}\right)\left(1-\dfrac{2}{4.5}\right)...\left(1-\dfrac{2}{n\left(n+1\right)}\right)\)
\(\Leftrightarrow C=\left(\dfrac{2.3-2}{2.3}\right)\left(\dfrac{3.4-2}{3.4}\right)\left(\dfrac{4.5-2}{4.5}\right)...\left(\dfrac{n\left(n-1\right)-2}{n\left(n+1\right)}\right)\)
\(\Leftrightarrow C=\dfrac{4}{2.3}.\dfrac{10}{3.4}.\dfrac{18}{4.5}...\left(\dfrac{n\left(n-1\right)-2}{n\left(n+1\right)}\right)\)
\(\Leftrightarrow C=\dfrac{1.4}{2.3}.\dfrac{2.5}{3.4}.\dfrac{3.6}{4.5}...\left(\dfrac{\left(n-1\right)\left(n+2\right)}{n\left(n+1\right)}\right)\)
\(\Leftrightarrow C=\dfrac{1.4.2.5.3.6...\left(n-1\right)\left(n+2\right)}{2.3.3.4.4.5.n\left(n+1\right)}\)
\(\Leftrightarrow C=\dfrac{\left[1.2.3...\left(n-1\right)\right]\left[4.5.6\left(n+2\right)\right]}{\left(2.3.4...n\right)\left[3.4.5....\left(n+1\right)\right]}\)
\(\Leftrightarrow C=\dfrac{n+2}{3n}\)
Vì \(\dfrac{n+2}{3n}< \dfrac{2n+2}{3n}\)
\(\Leftrightarrow C< \dfrac{2n+2}{3n}\)
Vậy ...
Giải:
\(C=\left(1-\dfrac{2}{2.3}\right)\left(1-\dfrac{2}{3.4}\right)\left(1-\dfrac{2}{4.5}\right)...\left(1-\dfrac{2}{n\left(n+1\right)}\right)\)
Đk: \(n\ne0;n\ne-1\)
\(C=\left(1-\dfrac{2}{2.3}\right)\left(1-\dfrac{2}{3.4}\right)\left(1-\dfrac{2}{4.5}\right)...\left(1-\dfrac{2}{n\left(n+1\right)}\right)\)
\(\Leftrightarrow C=\left(\dfrac{2.3-2}{2.3}\right)\left(\dfrac{3.4-2}{3.4}\right)\left(\dfrac{4.5-2}{4.5}\right)...\left(\dfrac{n\left(n-1\right)-2}{n\left(n+1\right)}\right)\)
\(\Leftrightarrow C=\dfrac{4}{2.3}.\dfrac{10}{3.4}.\dfrac{18}{4.5}...\left(\dfrac{n\left(n-1\right)-2}{n\left(n+1\right)}\right)\)
\(\Leftrightarrow C=\dfrac{1.4}{2.3}.\dfrac{2.5}{3.4}.\dfrac{3.6}{4.5}...\left(\dfrac{\left(n-1\right)\left(n+2\right)}{n\left(n+1\right)}\right)\)
\(\Leftrightarrow C=\dfrac{1.4.2.5.3.6...\left(n-1\right)\left(n+2\right)}{2.3.3.4.4.5.n\left(n+1\right)}\)
\(\Leftrightarrow C=\dfrac{\left[1.2.3...\left(n-1\right)\right]\left[4.5.6\left(n+2\right)\right]}{\left(2.3.4...n\right)\left[3.4.5....\left(n+1\right)\right]}\)
\(\Leftrightarrow C=\dfrac{n+2}{3n}\)
Vì \(\dfrac{n+2}{3n}< \dfrac{2n+2}{3n}\)
\(\Leftrightarrow C< \dfrac{2n+2}{3n}\)
Vậy ...
Ta có:
\(\frac{1}{2.5}+\frac{1}{5.8}+\frac{1}{8.11}+...+\frac{1}{\left(3n+2\right).\left(3n+5\right)}\)
\(\Rightarrow\frac{1}{3}.\left(\frac{3}{2.5}+\frac{3}{5.8}+\frac{3}{8.11}+...+\frac{3}{\left(3n+2\right).\left(3n+5\right)}\right)\)
\(\Rightarrow\frac{1}{3}.\left(\frac{1}{2}-\frac{1}{5}+\frac{1}{5}-\frac{1}{8}+...+\frac{1}{3n+2}-\frac{1}{3n+5}\right)\)
\(\Rightarrow\frac{1}{3}.\left(\frac{1}{2}-\frac{1}{3n+5}\right)\)
\(\Rightarrow\frac{1}{6}-\frac{1}{9n+15}\)
a: \(=\dfrac{2^5\cdot2^{12}\cdot2^6}{2^{24}}=\dfrac{1}{2}\)
b: \(=\dfrac{12-15}{20}\cdot\left(\dfrac{10-6}{30}\right)^2\)
\(=\dfrac{-3}{20}\cdot\left(\dfrac{2}{15}\right)^2=\dfrac{-3}{20}\cdot\dfrac{4}{225}=-\dfrac{1}{375}\)
c: \(=3\cdot\dfrac{2}{5}+2:\dfrac{1}{4}=1.2+8=9.2\)
Câu 2:
\(B=\dfrac{5^{21}\cdot\left(2\cdot5-9\right)}{5^{20}}\cdot\dfrac{7^{15}\left(7+3\right)}{15\cdot7^{15}-95\cdot7^{14}}\)
\(=\dfrac{5\cdot1}{1}\cdot\dfrac{7^{15}\cdot10}{7^{14}\cdot\left(15\cdot7-95\right)}\)
\(=5\cdot\dfrac{7\cdot10}{105-95}=5\cdot7=35\)
Sửa đề
\(A=\dfrac{1}{2.5}+\dfrac{1}{5.8}+...+\dfrac{1}{\left(3n-1\right)\left(3n+2\right)}\)
\(A=\dfrac{1}{3}\left(\dfrac{1}{2}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{8}+...+\dfrac{1}{3n-1}-\dfrac{1}{3n+2}\right)\)
\(A=\dfrac{1}{3}\left(\dfrac{1}{2}-\dfrac{1}{3n+2}\right)\)
\(A=\dfrac{1}{3}\left(\dfrac{3n+2}{6n+4}-\dfrac{2}{6n+4}\right)=\dfrac{1}{3}\left(\dfrac{3n+2-2}{6n+4}\right)=\dfrac{1}{3}\left(\dfrac{3n}{6n+4}\right)=\dfrac{3n}{18n+12}=\dfrac{3n}{3\left(6n+4\right)}=\dfrac{n}{6n+4}\)