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A= 1/7 - 1/10 + 1/10 - 1/13 + 1/13 - 1/16 + .... + 1/97 - 1/100
A= 1/7 - 1/100
A= 93/700
\(A=\frac{3}{7.10}+\frac{3}{10.13}+\frac{3}{13.16}+...+\frac{3}{97.100}\)
\(A=\frac{1}{7}-\frac{1}{10}+\frac{1}{10}-\frac{1}{13}+\frac{1}{13}-\frac{1}{16}+...+\frac{1}{37}-\frac{1}{100}\)
\(A=\frac{1}{7}-\frac{1}{100}\)
\(A=\frac{93}{700}\)
\(D=1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{10}+...+\frac{1}{97}-\frac{1}{100}\)
\(D=1-\frac{1}{100}\)
\(D=\frac{99}{100}\)
\(D=\frac{2}{1\cdot4}+\frac{2}{4\cdot7}+\frac{2}{7\cdot10}+...+\frac{2}{97\cdot100}\)
\(D=\frac{2}{3}\left(\frac{3}{1\cdot4}+\frac{3}{4\cdot7}+\frac{3}{7\cdot10}+...+\frac{3}{97\cdot100}\right)\)
\(D=\frac{2}{3}\left(1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{10}+...+\frac{1}{97}-\frac{1}{100}\right)\)
\(D=\frac{2}{3}\left(1-\frac{1}{100}\right)\)
\(D=\frac{2}{3}\cdot\frac{99}{100}=\frac{33}{50}\)
S=1/1-1/4+1/4+1/7-1/7+1/10+...+1/100-1/103
S=1/1-1/103
S=102/103
Vì 102/103<1 nên S<1
3/(1×4)+3/(4×7)+3/(7×10)+3/(10×13)+3/(13×16)
=1-1/4+1/4-1/7+1/7-1/10+1/10-1/13+1/13-1/16
=1-1/16
=15/16
31 x 434 x 737 x 10310 x 13 = 1.3289876e+12
mik phải dùng máy tính chứ có sịp nhân mới trả lời đc
nhỉ ?????
1/1*4 + 1/4*7 + 1/7*10 + ... + 1/97*100
= 1/3(3/1*4 + 3/4*7 + 3/7*10 + ... + 3/97*100)
= 1/3(1 - 1/4 + 1/4 - 1/7 + 1/7 - 1/10 + .... + 1/97 - 1/100)
= 1/3(1 - 1/100)
= 1/3*99/100
= 33/100
a) \(\dfrac{2}{1\times4}+\dfrac{2}{4\times7}+\dfrac{2}{7\times10}+...+\dfrac{2}{97\times100}\)
\(=2.\left(\dfrac{1}{1\times4}+\dfrac{1}{4\times7}+\dfrac{1}{7\times10}+...+\dfrac{1}{97\times100}\right)\)
\(=2.\left(1-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{10}+...+\dfrac{1}{97}-\dfrac{1}{100}\right)\)
\(=2.\left(1-\dfrac{1}{100}\right)\)
\(=2.\dfrac{99}{100}\)
\(=\dfrac{99}{50}\)
_____
b) \(\dfrac{3}{1\times5}+\dfrac{3}{5\times9}+\dfrac{3}{9\times13}+...+\dfrac{3}{97\times101}\)
\(=3.\left(\dfrac{1}{1\times5}+\dfrac{1}{5\times9}+\dfrac{1}{9\times13}+...+\dfrac{1}{97\times101}\right)\)
\(=3.\left(1-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{9}+\dfrac{1}{9}-\dfrac{1}{13}+...+\dfrac{1}{97}-\dfrac{1}{101}\right)\)
\(=3.\left(1-\dfrac{1}{101}\right)\)
\(=3.\dfrac{100}{101}\)
\(=\dfrac{300}{101}\)
\(\frac{3}{1\cdot4}+\frac{3}{4\cdot7}+...+\frac{3}{97\cdot100}\)
\(=1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+...+\frac{1}{97}-\frac{1}{100}\)
\(=1-\frac{1}{100}\)
\(=\frac{99}{100}\)
E= 7/4x7 + 7/7x10 =7/10x13+...+ 7/301x304
\(=\frac{7}{3}\left(\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{10}+...+\frac{1}{301}-\frac{1}{304}\right)\)
\(=\frac{7}{3}\left(\frac{1}{4}-\frac{1}{304}\right)\)
\(=\frac{7}{3}\cdot\frac{75}{304}\)
\(=\frac{175}{304}\)
E = \(\frac{7}{4.7}+\frac{7}{7.10}+\frac{7}{10.13}+...+\frac{7}{301.304}\)
=\(\frac{7}{3}.\frac{7-4}{4.7}+\frac{7}{3}.\frac{10-7}{7.10}+\frac{7}{3}.\frac{13-10}{10.13}+...+\frac{7}{3}.\frac{304-301}{301.304}\)
= \(\frac{7}{3}.\left(\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{10}+\frac{1}{10}-\frac{1}{13}+...+\frac{1}{301}-\frac{1}{304}\right)\)=\(\frac{7}{3}.\left(\frac{1}{4}-\frac{1}{304}\right)=\frac{7}{3}.\frac{75}{304}=\frac{175}{304}\)
\(A=\dfrac{1}{4}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{10}+...+\dfrac{1}{97}-\dfrac{1}{100}=\dfrac{1}{4}-\dfrac{1}{100}=\dfrac{6}{25}\)
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