Tính tổng : S =1x3x5+3x5x7+5x7x9+...+21x23x25
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\(\dfrac{3}{1\cdot3\cdot5}+\dfrac{3}{3\cdot5\cdot7}+\dfrac{3}{5\cdot7\cdot9}+\cdot\cdot\cdot+\dfrac{3}{37\cdot39\cdot41}\)
\(=\dfrac{3}{4}\left(\dfrac{4}{1\cdot3\cdot5}+\dfrac{4}{3\cdot5\cdot7}+\dfrac{4}{5\cdot7\cdot9}+\cdot\cdot\cdot+\dfrac{4}{37\cdot39\cdot41}\right)\)
\(=\dfrac{3}{4}\left(\dfrac{1}{1\cdot3}-\dfrac{1}{3\cdot5}+\dfrac{1}{3\cdot5}-\dfrac{1}{5\cdot7}+\dfrac{1}{5\cdot7}-\dfrac{1}{7\cdot9}+\cdot\cdot\cdot+\dfrac{1}{37\cdot39}-\dfrac{1}{39\cdot41}\right)\)
\(=\dfrac{3}{4}\left(\dfrac{1}{3}-\dfrac{1}{39\cdot41}\right)\)
\(=\dfrac{3}{4}\cdot\dfrac{532}{1599}=\dfrac{133}{533}\)
#Ayumu
\(\frac{4}{1.3.5}+\frac{4}{3.5.7}+\frac{4}{5.7.9}+\frac{4}{7.9.11}+\frac{4}{9.11.13}\)
= \(\frac{1}{1.3.5}+\frac{1}{3.5.7}+\frac{1}{5.7.9}+\frac{1}{7.9.11}+\frac{1}{9.11.13}\)
= \(\frac{1}{1.3}-\frac{1}{3.5}+\frac{1}{3.5}-\frac{1}{5.7}+...+\frac{1}{9.11}-\frac{1}{11.13}\)
= \(\frac{1}{1.3}-\frac{1}{11.13}\)
= \(\frac{140}{429}\)
~~~
Không chắc chắn lắm nhé :3
#Sunrise
nhớ cho k nhé !
4/1x3x5 = 1/1x3 - 1/3x5
4/3x5x7 = 1/3x5 - 1/5x7
.............
A = 1/1x3 - 1/11x13
1/1x3x5 = 1/4 x (1/1x3 - 1/3x5)
1/3x5x7 = 1/4 x (1/3x5 - 1/5x7)
..........
B = 1/4 x (1/1x3 - 1/11x13)
\(A=9\left(\dfrac{4}{1\cdot3\cdot5}+\dfrac{4}{3\cdot5\cdot7}+...+\dfrac{4}{25\cdot27\cdot29}\right)\)
\(=9\left(\dfrac{1}{1\cdot3}-\dfrac{1}{3\cdot5}+\dfrac{1}{3\cdot5}-\dfrac{1}{5\cdot7}+...+\dfrac{1}{25\cdot27}-\dfrac{1}{27\cdot29}\right)\)
\(=9\left(\dfrac{1}{3}-\dfrac{1}{783}\right)=\dfrac{260}{87}\)
\(\frac{4}{1x3x5}+\frac{4}{3x5x7}+...+\frac{4}{9x11x13}\)
\(=\frac{1}{1x3}-\frac{1}{3x5}+\frac{1}{3x5}-...+\frac{1}{9x11}-\frac{1}{11x13}\)
\(=\frac{1}{3}-\frac{1}{143}\)
\(=\frac{140}{429}\)