\(1^2+2^2+3^2+4^2+........+19^2+20^2\)
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Cái này chỉ cần bỏ ngoặc ghép cặp lại rồi tính là được mà, mỗi cặp = 1
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\(\dfrac{\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+...+\dfrac{1}{20}}{\dfrac{19}{1}+\dfrac{18}{2}+\dfrac{17}{3}+....+\dfrac{1}{19}}\)
\(=\dfrac{\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+...+\dfrac{1}{20}}{1+\left(\dfrac{18}{2}+1\right)+\left(\dfrac{17}{3}+1\right)+\left(\dfrac{1}{19}+1\right)}\)
\(=\dfrac{\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{20}}{1+\dfrac{20}{2}+\dfrac{20}{3}+...+\dfrac{20}{19}}\)
\(=\dfrac{\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{20}}{20.\left(\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{19}+\dfrac{1}{20}\right)}\)
\(=\dfrac{1}{20}\)
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\(\frac{2}{1.2}+\frac{2}{2.3}+\frac{2}{3.4}+....+\frac{2}{18.19}+\frac{2}{19.20}\)
\(=2\cdot\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{18}-\frac{1}{19}+\frac{1}{19}-\frac{1}{20}\right)\)
\(=2\cdot\left(1-\frac{1}{20}\right)\)
\(=2\cdot\frac{19}{20}=\frac{19}{10}\)
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câu 1:
2 + 2^2 + 2^3 + ... + 2^20 = 2( 1 + 2 + 2^2 +... + 2^19) chia hết cho 2
câu 2
2 + 2^2 + 2^3 + 2^4 +... + 2^19 + 2^20
= ( 2 + 2^2) + ( 2^3 + 2^4) + ....+ ( 2^19 + 2^20)
= 2( 1 + 2 ) + 2^3( 1+3) +...+ 2^19(1+2)
= 2. 3 + 2^3 . 3 +...+2^19.3
= 3.(2+2^3+2^5+....+2^19) chia hết cho 3
\(a.2+2^2+2^3+...+2^{19}\)\(+2^{20}\)
Ta có: \(2⋮2,2^2,2^3⋮2,..2^{19}⋮2,2^{20}⋮2\)
\(\Rightarrow2+2^2+2^3+...+2^{19}+2^{20}⋮2\)
b.Giống trên
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1.(2 - 1)+2.(3 - 1)+3.(4 - 1)+....+20.(20 + 1 - 1)=[(1.2 +2.3 + 3.4 +4.5 + ....+20.(20+1)] - (1 + 2 +3 + ... +20)=\(\frac{20.\left(20+1\right).\left(20.2+1\right)}{6}\)
=2870
Đặt\(A=1^2+2^2+3^2+...+20^2\)
Ta có \(A=1\cdot\left(2-1\right)+2\cdot\left(3-1\right)+3\cdot\left(4-1\right)+...+20\cdot\left(21-1\right)\)
\(A=\left(1\cdot2+2\cdot3+...+20\cdot21\right)-\left(1+2+3+...+20\right)\)
\(A=B-C\)(với \(B=\left(1\cdot2+2\cdot3+...+20\cdot21\right);C=\left(1+2+3+...+20\right)\)
Dễ nhận thấy \(C=1+2+3+...+20=\frac{20\cdot21}{2}=10\cdot21=210\)
Xét \(3B=1\cdot2\cdot3+2\cdot3\cdot\left(4-1\right)+3\cdot4\cdot\left(5-2\right)+...+20\cdot21\cdot\left(22-19\right)\)
\(3B=\left(1\cdot2\cdot3+2\cdot3\cdot4+...+20\cdot21\cdot22\right)-\left(1\cdot2\cdot3+2\cdot3\cdot4+...+19\cdot20\cdot21\right)\)
\(3B=20\cdot21\cdot22\Leftrightarrow B=20\cdot7\cdot22=3080\)
Vậy \(A=B-C=3080-210=2870\)
Nhận xét: Phương pháp giải
Tính A bằng cách đưa về những dãy số đã biết cách tính
Tính B bằng cách khử liên tiếp: số hạng sau sẽ khử số hạng liền trước.
Chúc bạn học tốt!
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1) \(\dfrac{1}{2}+\dfrac{13}{19}-\dfrac{4}{9}+\dfrac{6}{19}+\dfrac{5}{18}\)
\(=\dfrac{1}{2}+\left(\dfrac{13}{19}+\dfrac{6}{19}\right)-\dfrac{4}{9}+\dfrac{5}{18}\)
\(=\dfrac{3}{2}-\dfrac{4}{9}+\dfrac{5}{18}\)
\(=\dfrac{19}{18}+\dfrac{5}{18}\)
\(=\dfrac{24}{18}\)
\(=\dfrac{4}{3}\)
2) \(\dfrac{-20}{23}+\dfrac{2}{3}-\dfrac{3}{23}+\dfrac{2}{5}+\dfrac{7}{15}\)
\(=\left(-\dfrac{20}{23}-\dfrac{3}{23}\right)+\dfrac{2}{3}+\dfrac{2}{5}+\dfrac{7}{15}\)
\(=-1+\dfrac{2}{3}+\dfrac{2}{5}+\dfrac{7}{15}\)
\(=-\dfrac{1}{3}+\dfrac{2}{5}+\dfrac{7}{15}\)
\(=\dfrac{1}{15}+\dfrac{7}{15}\)
\(=\dfrac{8}{15}\)
3) \(\dfrac{4}{3}+\dfrac{-11}{31}+\dfrac{3}{10}-\dfrac{20}{31}-\dfrac{2}{5}\)
\(=\left(\dfrac{-11}{31}-\dfrac{20}{31}\right)+\dfrac{4}{3}+\dfrac{3}{10}-\dfrac{2}{5}\)
\(=-1+\dfrac{4}{3}+\dfrac{3}{10}-\dfrac{2}{5}\)
\(=\dfrac{1}{3}+\dfrac{3}{10}-\dfrac{2}{5}\)
\(=\dfrac{1}{3}-\dfrac{1}{10}\)
\(=\dfrac{7}{30}\)
4) \(\dfrac{5}{7}.\dfrac{5}{11}+\dfrac{5}{7}.\dfrac{2}{11}-\dfrac{5}{7}.\dfrac{14}{11}\)
\(=\dfrac{5}{7}.\left(\dfrac{5}{11}+\dfrac{2}{11}-\dfrac{14}{11}\right)\)
\(=\dfrac{5}{7}.-\dfrac{7}{11}\)
\(=-\dfrac{35}{77}\)
\(=-\dfrac{5}{11}\)