Chú ý: \(a^2-1=\left(a-1\right)\left(a+1\right)\)

Áp dụng:

\(A=\frac{2.4}{3^2}.\frac{3.5}{4^2}.\frac{4.6}{5^2}...\frac{49.51}{50^2}=\frac{2.3.4^2.5^2...49^2.50.51}{3^2.4^2.5^2...50^2}=\frac{2.51}{3.50}=\frac{51}{75}\)

\(S=\frac{1}{4}+\left(\frac{1}{3^2}+\frac{1}{4^2}+..+\frac{1}{50^2}\right)\)

Xét \(A=\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{50^2}\)

\(A< \frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{49.50}\)

\(A< \frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{49}-\frac{1}{50}\)

\(A< \frac{1}{2}-\frac{1}{50}< \frac{1}{2}\)

\(=>A< \frac{1}{2}\)

=>\(S=\frac{1}{4}+A< \frac{1}{4}+\frac{1}{2}=\frac{3}{4}\)

vậy S<3/4

\(=\left(1-\frac{1}{3^2}\right)\left(1-\frac{1}{4^2}\right)\left(1-\frac{1}{5^2}\right)...\left(1-\frac{1}{50^2}\right)\)

\(=\frac{8}{3\cdot3}\cdot\frac{15}{4\cdot4}\cdot\frac{24}{5\cdot5}\cdot....\cdot\frac{2499}{50\cdot50}\)

\(=\frac{\left(2\cdot4\right)\left(3\cdot5\right)\left(4\cdot6\right)...\left(49\cdot51\right)}{\left(3\cdot3\right)\left(4\cdot4\right)\left(5\cdot5\right)...\left(50\cdot50\right)}\)

\(=\frac{\left(2\cdot3\cdot4\cdot...\cdot49\right)\left(4\cdot5\cdot6\cdot...\cdot51\right)}{\left(3\cdot4\cdot5\cdot...\cdot50\right)\left(3\cdot4\cdot5\cdot...\cdot50\right)}\)

\(=\frac{2\cdot51}{50\cdot3}\)

Đặt:  \(A=1+2^1+2^2+2^3+...+2^{50}\)

\(\Rightarrow2A=2^1+2^2+2^3+2^4+...+2^{51}\)

\(\Rightarrow2A-A=\left(2^1+2^2+2^3+2^4+...+2^{51}\right)-\left(1+2^1+2^2+2^3+...+2^{50}\right)\)

\(\Rightarrow A=2^{51}-1\)

1 + 2¹ + 2² + 2³ + ... + 2⁵⁰

Ta có : Đặt A = 1 + 2¹ + 2² + 2³ + ... + 2⁵⁰

                2A = 2(1 + 2¹ + 2² + 2³ + ... + 2⁵⁰)

                 2A = 2 + 2² + 2³  + ... + 2⁵¹

              2A - A = ( 2 + 2² + 2³  + ... + 2⁵¹) - (1 + 2¹ + 2² + 2³ + ... + 2⁵⁰)

                    A = 2⁵¹ - 1

1² + 2² + 3² + .... + 50²

= 1(2 - 1) + 2.(3 - 1) + 3.(4 - 1) + ..... + 50(51 - 1)

= 1.2 - 1 + 2.3 - 2 + 3.4 - 3 + ..... + 50.51 - 50

= (1.2 + 2.3 + ... + 50.51) - (1 + 2 + ... + 50)

\(=\frac{50.51.52}{3}-\frac{50.51}{2}\)

\(=44200-1275\)

= 42925

\(2^1+2^2+2^3+...+2^{50}\\ =\left(2-1\right)\left(2^1+2^2+2^3+...+2^{50}\right)\\ =2^2-2+2^3-2^2+2^4-2^3+...+2^{51}-2^{50}\\ =2^{51}-2\)

ta đặc \(2^1+2^2+2^3+...+2^{50}\) là \(B\)

\(\Rightarrow2B=2\left(2^1+2^2+2^3+...+2^{50}\right)\)

\(2B=2^2+2^3+2^4+...+2^{51}\)

ta có : \(2B-B=B=\left(2^2+2^3+2^4+...+2^{51}\right)-\left(2^1+2^2+2^3+...+2^{50}\right)\)

\(B=2^{51}-2^1=2^{51}-2\)

vậy \(2^1+2^2+2^3+...+2^{50}=2^{51}-2\)

a, \(-1\dfrac{2}{3}+\dfrac{3}{4}-\dfrac{1}{2}+2\dfrac{1}{6}\\ =-\dfrac{5}{3}+\dfrac{3}{4}-\dfrac{1}{2}+\dfrac{13}{6}\\ =\dfrac{-5.4+3.3-1.6+13.2}{12}=\dfrac{9}{12}=\dfrac{3}{4}\)

b, \(\dfrac{11}{50}\left(-17\dfrac{1}{2}\right)-\dfrac{11}{50}.82\dfrac{1}{2}\\ =\dfrac{11}{50}.\left(-17\dfrac{1}{2}-82\dfrac{1}{2}\right)=\dfrac{11}{50}.\left(-100\right)=-22\)

a) \(-1\dfrac{2}{3}\) + \(\dfrac{3}{4}\) \(-\) \(\dfrac{1}{2}\) + \(2\dfrac{1}{6}\)

=\(-\dfrac{5}{3}\) + \(\dfrac{3}{4}\) \(-\) \(\dfrac{1}{2}\) + \(\dfrac{13}{6}\)\()\)

=\(-\) \(\dfrac{20}{12}\) + \(\dfrac{9}{12}\) \(-\) \(\dfrac{6}{12}\) + \(\dfrac{26}{12}\)

= \((\)\(\dfrac{-20}{12}\) + \(\dfrac{26}{12}\) \()\) + \((\) \(\dfrac{9}{12}\) \(-\) \(\dfrac{6}{12}\) \()\)

= \(\dfrac{1}{2}\) + \(\dfrac{1}{4}\)

= \(\dfrac{3}{4}\)

b)\(\dfrac{11}{50}\) \((\) \(-17\dfrac{1}{2}\) \()\) \(-\) \(\dfrac{11}{50}\) .\(82\dfrac{1}{2}\)

= \(\dfrac{11}{50}\) . \(-\dfrac{35}{2}\) \(-\) \(\dfrac{11}{50}\) . \(\dfrac{165}{2}\)

= \(\dfrac{11}{50}\). \((\) \(-\dfrac{35}{2}\) \(-\) \(\dfrac{165}{2}\) \()\)

=\(\dfrac{11}{50}\). \(-\)\(100\)

= \(-22\)

Chúc bạn học thật tốt nha !