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Đặt S = 1 x 2 + 2 x 3 + ..... + 9 x 10
3S = 1 x 2 x (3-0) + 2 x 3 x (4 - 1) + ....... + 9 x 10 x (11 - 8)
3S = 1 x 2 x 3 + 2 x 3 x 4 - 1 x 2 x 3 + ......... + 9 x 10 x 11 - 8 x 9 x 10
3S = ( 1 x 2 x 3 -1 x 2 x 3) + ( 2 x 3 x4 - 2 x 3 x 4) + ..... + (8 x 9 x 10 - 8 x 9 x 10) + 9 x 10 x 11
3S = 9 x 10 x 11 = 990
S = 990 : 3 = 330
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Mình làm mẫu 1 bài rùi bạn tự giải những bài còn lại nha
1, 7A = 7+7^2+7^3+....+7^2008
6A = 7A - A = (7+7^2+7^3+....+7^2008)-(1+7+7^2+....+7^2007) = 7^2008-1
=> A = (7^2008-1)/6
Tk mk nha
\(A=1+7+7^2+7^3+...+7^{2007}\)
\(\Rightarrow7A=7+7^2+7^3+7^4+...+7^{2008}\)
\(\Rightarrow7A-A=\left(7+7^2+7^3+...+7^{2008}\right)-\left(1+7+7^2+...+7^{2007}\right)\)
\(\Rightarrow6A=7^{2008}-1\)
\(\Rightarrow A=\frac{7^{2008}-1}{6}\)
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Bài 1 a, \(17.25\)
\(=\left(1+16\right).25\)
\(=1.25+16.25\)
\(=25+4.4.25\)
\(=25+4.100\)
\(=25+400\)
\(=425\)
b, Bạn kia làm rồi nhé!
c, \(29.31+66.69+31.37\)
\(=31.\left(29+37\right)+66.99\)
\(=31.66+66.69\)
\(=66\left(31+69\right)\)
\(=66.100\)
\(6600\)
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a: \(\dfrac{1}{2}+\dfrac{-3}{8}+\dfrac{5}{9}\)
\(=\dfrac{36}{72}-\dfrac{27}{72}+\dfrac{40}{72}\)
\(=\dfrac{49}{72}\)
b: \(\dfrac{13}{-30}+\dfrac{17}{45}+\dfrac{-7}{18}\)
\(=\dfrac{-13}{30}+\dfrac{17}{45}+\dfrac{-7}{18}\)
\(=\dfrac{-39+34-35}{90}=\dfrac{-40}{90}=-\dfrac{4}{9}\)
c: \(-\dfrac{5}{19}+\dfrac{7}{4}-\left(-\dfrac{2}{3}\right)+\dfrac{5}{6}\)
\(=-\dfrac{5}{19}+\dfrac{7}{4}+\dfrac{2}{3}+\dfrac{5}{6}\)
\(=\dfrac{-5}{19}+\dfrac{21+8+10}{12}\)
\(=-\dfrac{5}{19}+\dfrac{39}{12}=\dfrac{-5}{19}+\dfrac{13}{4}=\dfrac{227}{76}\)
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a)\(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{9.10}\)
= \(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{3}{4}+...+\frac{1}{9}-\frac{1}{10}\)
= \(1+\left(\frac{-1}{2}+\frac{1}{2}\right)+\left(\frac{-1}{3}+\frac{1}{3}\right)+...+\left(\frac{-1}{9}+\frac{1}{9}\right)-\frac{1}{10}\)
= \(1-\frac{1}{10}\)
=\(\frac{9}{10}\)
b)\(\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+\frac{2}{7.9}+\frac{2}{9.11}\)
= \(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+\frac{1}{9}-\frac{1}{11}\)
=\(1+\left(\frac{-1}{3}+\frac{1}{3}\right)+\left(\frac{-1}{5}+\frac{1}{5}\right)+\left(\frac{-1}{7}+\frac{1}{7}\right)+\left(\frac{-1}{9}+\frac{1}{9}\right)-\frac{1}{11}\)
=\(1-\frac{1}{11}\)
= \(\frac{10}{11}\)
c) đặt A=\(\frac{3}{1.3}+\frac{3}{3.5}+\frac{3}{5.7}+\frac{3}{7.9}+\frac{3}{9.11}\)
\(\frac{1}{3}A\)=\(\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+\frac{1}{7.9}+\frac{1}{9.11}\)
\(\frac{2}{3}A\)=\(\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+\frac{2}{7.9}+\frac{2}{9.11}\)
\(\frac{2}{3}A\)=\(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+\frac{1}{9}-\frac{1}{11}\)
\(\frac{2}{3}A\)=\(1+\left(\frac{-1}{3}+\frac{1}{3}\right)+\left(\frac{-1}{5}+\frac{1}{5}\right)+\left(\frac{-1}{7}+\frac{1}{7}\right)+\left(\frac{-1}{9}+\frac{1}{9}\right)-\frac{1}{11}\)
\(\frac{2}{3}A\)=\(\frac{10}{11}\)
A= \(\frac{10}{11}:\frac{2}{3}\)
A= \(\frac{10}{11}.\frac{3}{2}\)=\(\frac{15}{11}\)
d) giả tương tự câu c kết quả \(\frac{25}{11}\)
tổng đặc biệt đó bạn
\(\frac{1}{1\times2}+\frac{1}{2\times3}+\frac{1}{3\times4}+...+\frac{1}{9\times10}\)
\(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{9}-\frac{1}{10}\)
\(1-\frac{1}{10}=\frac{9}{10}\)
những câu sau cũng áp dụng như vậy nhé