C=1x3x5-3x5x7+5x7x9-7x9x11+...+97x99x101
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\(\frac{4}{1x3x5}+\frac{4}{3x5x7}+...+\frac{4}{9x11x13}\)
\(=\frac{1}{1x3}-\frac{1}{3x5}+\frac{1}{3x5}-...+\frac{1}{9x11}-\frac{1}{11x13}\)
\(=\frac{1}{3}-\frac{1}{143}\)
\(=\frac{140}{429}\)
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\(\frac{4}{1.3.5}+\frac{4}{3.5.7}+\frac{4}{5.7.9}+\frac{4}{7.9.11}+\frac{4}{9.11.13}\)
= \(\frac{1}{1.3.5}+\frac{1}{3.5.7}+\frac{1}{5.7.9}+\frac{1}{7.9.11}+\frac{1}{9.11.13}\)
= \(\frac{1}{1.3}-\frac{1}{3.5}+\frac{1}{3.5}-\frac{1}{5.7}+...+\frac{1}{9.11}-\frac{1}{11.13}\)
= \(\frac{1}{1.3}-\frac{1}{11.13}\)
= \(\frac{140}{429}\)
~~~
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)
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\(A=\frac{2.6.10+6.10.14+...+1994.1998.2002}{1.3.5+3.5.7+...+97.99.101}\)
\(=\frac{1.2^3.3.5+3.2^3.5.7+....+2^3.97.99.101}{1.3.5+3.5.7+...+97.99.101}\)
\(=\frac{2^3\left(1.3.5+3.5.7+...+97.99.101\right)}{1.3.5+3.5.7+...+97.99.101}\)
\(=8\)
Vậy A = 8