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a) \(A=\frac{2}{1.4}+\frac{2}{4.7}+\frac{2}{7.10}+...+\frac{2}{31.34}\)
\(A=\frac{2}{3}.\left(1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{10}+...+\frac{1}{31}-\frac{1}{34}\right)\)
\(A=\frac{2}{3}.\left(1-\frac{1}{34}\right)\)
\(A=\frac{2}{3}\cdot\frac{33}{34}=\frac{11}{17}\)
b) \(B=\frac{3}{1}+\frac{3}{3}+\frac{3}{6}+...+\frac{3}{210}\)
\(B=\frac{6}{2}+\frac{6}{6}+\frac{6}{12}+...+\frac{6}{420}\) ( 3/1 = 6/2; 6/6=3/3;..)
\(B=\frac{6}{1.2}+\frac{6}{2.3}+\frac{6}{3.4}+...+\frac{6}{20.21}\)
\(B=6.\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{20}-\frac{1}{21}\right)\)
\(B=6.\left(1-\frac{1}{21}\right)=6\cdot\frac{20}{21}=\frac{40}{7}\)
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\(A = 1 + 4 + 4^2 + ... + 4\)\(20\)
\(4A = 4 + 4^2 + 4^3 + ...+ 4\)\(21\)
\(4A - A = ( 4+ 4^2 + 4^3 + ... + 4\)\(21\)\()\)\(- ( 1 + 4 + 4^2 + ... + 4\)\(20\) \()\)
\(3A = 2\)\(21\) \(- 1\)
\(\Leftrightarrow\)\(3A + 1 = 2\)\(21\)\(= ( 2^3)^7\)\(= 8^7\)
\(Ta có : 8^7 < 63^7 \)
\(Nên 3A + 1 < 63^7\)
Vì A= 4^0 + 4^1 + 4^2+ 4^3+....+4^20
Suy ra: 4A= 4^1+4^2+4^3+4^4+......+ 4^21
Suy ra:4A-A= 4^21 - 4^0
Suy ra: 3A = 4^21-1
Suy ra: A= (4^21-1) : 3
Suy ra: 3A+1= 3. [ ( 4^21-1) : 3] +1
Suy ra: 3A+1 = ( 4^21-1)+1
Suy ra: 3A + 1 = 4^21= (4^3)^7=64^7
Vì 64 > 63; 7=7
Suy ra: 64^7 > 63^7 hay 3A+1 > 63^7
![](https://rs.olm.vn/images/avt/0.png?1311)
Ta có: 4;42;43;...;429;430 chia hết cho 4
Nên 4+42+43+...+429+430 chia hết cho 4
Nên A=1+4+42+...+429+430 không chia hết cho 4( vì 1 không chia hết cho 4)
Vậy A không chia hết cho 4
![](https://rs.olm.vn/images/avt/0.png?1311)
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\(A=\frac{3}{1.4}+\frac{3}{4.7}+...+\frac{3}{97.100}\)
\(\Rightarrow A=1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+...+\frac{1}{97}-\frac{1}{100}\)
\(\Rightarrow A=1-\frac{1}{100}=\frac{99}{100}\)
\(A=\frac{3}{1.4}+\frac{3}{4.7}+\frac{3}{7.10}+...+\frac{3}{97.100}\)
\(A=1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{10}+...+\frac{1}{97}-\frac{1}{100}\)
\(A=1-\frac{1}{100}\)
\(A=\frac{99}{100}\)
Vậy\(A=\frac{99}{100}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(A=4+4^2+4^3+...+4^{2018}\)
\(4A=4^2+4^3+4^4+...+4^{2019}\)
\(3A=4^{2019}-4\)
\(A=\frac{4^{2019}-4}{3}\)
Đề là tính hả chả ghi đề j -_-"
\(4A=4^2+4^3+...+4^{2019}\)
\(4A-A=\left(4^2+4^3+...+4^{2019}\right)-\left(4+4^2+...+4^{2018}\right)\)
\(3A=4^{2019}-4\)
\(A=\frac{4^{2019}-4}{3}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(A=4^1+4^2+4^3+...+4^{50}\)
\(4A=4^2+4^3+4^4+...+4^{51}\)
\(4A-A=\left(4^2+4^3+4^4+...+4^{51}\right)-\left(4^1+4^2+4^3+...+4^{50}\right)\)
\(3A=4^{51}-4\)
\(A=\frac{4^{51}-4}{3}\)
Đề tự vt đi t giải luôn
\(4A=4^2+4^3+...+4^{51}\)
\(4A-A=\left(4^2+4^3+...+4^{51}\right)-\left(4+4^2+...+4^{50}\right)\)
\(3A=4^{51}-4\)
\(A=\frac{4^{51}-4}{3}\)
A=(4^60-1):3