Chứng minh rằng:A=36/1x3x5+36/3x5x7+36/5x7x9+....+36/25x27x29
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A = 1/1x2x3 + 1/2x3x4 + 1/3x4x5 +...+ 1/18x19x20 < 1/4B = 36/1x3x5 + 36/3x5x7 + 36/5x7x9 +...+ 36/25x27x29 < 3C = 1/2^2 + 1/3^2 + 1/4^2 +...+ 1/100^2 < 1D = 1/101 + 1/102 + 1/103 +...+ 1/999 + 1/200 > 7/12
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2A=\(\frac{2}{1.2.3}\)+\(\frac{2}{2.3.4}\)+...+\(\frac{2}{18.19.20}\)
=1/1.2-1/2.3+1/2.3-1/3.4+...+1/18.19-1/19.20
=1/2-1/19.20
A=1/4-1/19.20.2
vậy A<1/4
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Bạn nào giải giúp mình bài này với:
D= 36/1x3x5+36/3x5x7+...+36/25x27x29
Mình cảm ơn nhiều!
C=1x3x5-3x5x7+5x7x9-7x9x11+...+97x99x101
Tính tổng : S =1x3x5+3x5x7+5x7x9+...+21x23x25
1x3x5+3x5x7+5x7x9+...+97x99x101
Giúp với ạ
tính nhanh :
4/1x3x5 + 4/3x5x7 + 4/5x7x9 + 4/7x9x11 + 4/9x11x13 .
4/1x3x5 + 4/3x5x7 + 4/5x7x9 + 4/7x9x11 + 4/9x11x13
\(\frac{4}{1x3x5}+\frac{4}{3x5x7}+...+\frac{4}{9x11x13}\)
\(=\frac{1}{1x3}-\frac{1}{3x5}+\frac{1}{3x5}-...+\frac{1}{9x11}-\frac{1}{11x13}\)
\(=\frac{1}{3}-\frac{1}{143}\)
\(=\frac{140}{429}\)
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4/1x3x5+4/3x5x7+4/5x7x9+4/7x9x11+4/9x11x13
=1/1x3-1/3x5+1/3x5-1/5x7+1/5x7-1/7x9+1/7x9-1/9x11+1/9x11-1/11x13
=1/1x3-1/11x13
=1/3-1/143=140/429
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Xem thêm câu trả lời
\(\dfrac{4}{1x3x5}+\dfrac{4}{3x5x7}+\dfrac{4}{5x7x9}+\dfrac{4}{9x11x13}\)
\(\dfrac{4}{1\times3\times5}+\dfrac{4}{3\times5\times7}+\dfrac{4}{5\times7\times9}+\dfrac{4}{7\times9\times11}\)
=\(\dfrac{5-1}{1\times3\times5}+\dfrac{7-3}{3\times5\times7}+\dfrac{9-5}{5\times7\times9}+\dfrac{11-7}{7\times9\times11}\)
=\(\dfrac{1}{1\times3}-\dfrac{1}{3\times5}+\dfrac{1}{3\times5}-\dfrac{1}{5\times7}+...+\dfrac{1}{9\times11}-\dfrac{1}{11\times13}\)
=\(\dfrac{1}{3}-\dfrac{1}{143}=\dfrac{140}{429}\)
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tính nhanh 4/1x3x5+4/3x5x7+4/5x7x9+4/7x9x11+4/9x11x13
