10 Rút gọn:
a) A= 1+2+22+23+24+...+249+250
b) B= \(\dfrac{1}{2}+(\dfrac{1}{2})^2+(\dfrac{1}{2})^3+(\dfrac{1}{2})^4+(\dfrac{1}{2})^5+...+(\dfrac{1}{2})^{99}+(\dfrac{1}{2})^{100}\)
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a: \(\dfrac{4^5\cdot9^4-2\cdot6^9}{2^{10}\cdot3^8+6^8\cdot20}\)
\(=\dfrac{2^{10}\cdot3^8-2\cdot2^9\cdot3^9}{2^{10}\cdot3^8+2^8\cdot3^8\cdot2^2\cdot5}\)
\(=\dfrac{2^{10}\cdot3^8-2^{10}\cdot3^9}{2^{10}\cdot3^8+2^{10}\cdot3^8\cdot5}\)
\(=\dfrac{2^{10}\cdot3^8\left(1-3\right)}{2^{10}\cdot3^8\left(1+5\right)}=\dfrac{-2}{6}=-\dfrac{1}{3}\)
b, B = \(\dfrac{1}{2}\) - \(\dfrac{1}{2^2}\) + \(\dfrac{1}{2^3}\) - \(\dfrac{1}{2^4}\)+.....+ \(\dfrac{1}{2^{99}}\) - \(\dfrac{1}{2^{100}}\)
2 \(\times\) B = 1 + \(\dfrac{1}{2}\) + \(\dfrac{1}{2^2}\) - \(\dfrac{1}{2^3}\) + \(\dfrac{1}{2^4}\)-.......-\(\dfrac{1}{2^{99}}\)
2 \(\times\) B + B = 1 - \(\dfrac{1}{2^{100}}\)
3B = ( 1 - \(\dfrac{1}{2^{100}}\))
B = ( 1 - \(\dfrac{1}{2^{100}}\)) : 3
A = 1 + \(\dfrac{1}{3}\) + \(\dfrac{1}{3^2}\)+ \(\dfrac{1}{3^3}\)+......+ \(\dfrac{1}{3^{n-1}}\) + \(\dfrac{1}{3^n}\)
A\(\times\) 3 = 3 + 1 + \(\dfrac{1}{3}\) + \(\dfrac{1}{3^2}\) + \(\dfrac{1}{3^2}\)+....+ \(\dfrac{1}{3^{n-1}}\)
A \(\times\) 3 - A = 3 - \(\dfrac{1}{3^n}\)
2A = 3 - \(\dfrac{1}{3^n}\)
A = ( 3 - \(\dfrac{1}{3^n}\)) : 2
Với n\(\in N\)* có: \(\dfrac{1}{\left(n+1\right)\sqrt{n}+n\sqrt{n+1}}=\dfrac{1}{\sqrt{n\left(n+1\right)}\left(\sqrt{n+1}+\sqrt{n}\right)}\)\(=\dfrac{\sqrt{n+1}-\sqrt{n}}{\sqrt{n\left(n+1\right)}\left(n+1-n\right)}=\dfrac{\sqrt{n+1}-\sqrt{n}}{\sqrt{n\left(n+1\right)}}\)\(=\dfrac{1}{\sqrt{n}}-\dfrac{1}{\sqrt{n+1}}\)
\(\Rightarrow\)\(\dfrac{1}{\left(n+1\right)\sqrt{n}+n\sqrt{n+1}}=\dfrac{1}{\sqrt{n}}-\dfrac{1}{\sqrt{n+1}}\) (*)
a) Áp dụng (*) vào T
\(\Rightarrow T=1-\dfrac{1}{\sqrt{2}}+\dfrac{1}{\sqrt{2}}-\dfrac{1}{\sqrt{3}}+...+\dfrac{1}{\sqrt{99}}-\dfrac{1}{\sqrt{100}}\)\(=1-\dfrac{1}{10}=\dfrac{9}{10}\)
b) Có \(VT=1-\dfrac{1}{\sqrt{2}}+\dfrac{1}{\sqrt{2}}-\dfrac{1}{\sqrt{3}}+...+\dfrac{1}{\sqrt{n}}-\dfrac{1}{\sqrt{n+1}}\)\(=1-\dfrac{1}{\sqrt{n+1}}=\dfrac{4}{5}\)
\(\Leftrightarrow\sqrt{n+1}=5\Leftrightarrow n=24\) (tm)
Vậy n=24.
\(A=\dfrac{1}{\sqrt{1}+\sqrt{2}}+\dfrac{1}{\sqrt{2}+\sqrt{3}}+...+\dfrac{1}{\sqrt{99}+\sqrt{100}}\)
\(=\dfrac{\sqrt{2}-\sqrt{1}}{\left(\sqrt{1}+\sqrt{2}\right)\left(\sqrt{2}-\sqrt{1}\right)}+\dfrac{\sqrt{3}-\sqrt{2}}{\left(\sqrt{3}-\sqrt{2}\right)\left(\sqrt{3}+\sqrt{2}\right)}+...+\dfrac{\sqrt{100}-\sqrt{99}}{\left(\sqrt{100}-\sqrt{99}\right)\left(\sqrt{100}+\sqrt{99}\right)}\)
\(=\sqrt{2}-\sqrt{1}+\sqrt{3}-\sqrt{2}+...+\sqrt{100}-\sqrt{99}=\sqrt{100}-\sqrt{1}=10-1=9\)
cả 2 ý bạn trục căn thức ở mấu là xong nhé:
vd: \(\dfrac{1}{\sqrt{1}+\sqrt{2}}=\dfrac{\sqrt{1}-\sqrt{2}}{-1}\). Rồi tương tự như vậy
b.ta chia B thành 10 nhóm mỗi nhóm có 6 hạng tử \(B=\left(2+2^2+2^3+2^4+2^5+2^6\right)+....+\left(2^{55}+2^{56}+2^{57}+2^{58}+2^{59}+2^{60}\right)\)
\(B\text{=}2\left(1+2+2^2+2^3+2^4+2^5\right)+...+2^{55}\left(1+2+2^2+2^3+2^4+2^5\right)\)
\(B\text{=}2.63+...+2^{56}.63\)
\(\Rightarrow B⋮63\)
\(\Rightarrow B⋮21\)
a) \(A=2A-A\)
\(=2\left(\dfrac{1}{2}+\dfrac{1}{2^2}+...+\dfrac{1}{2^{2022}}\right)-\left(\dfrac{1}{2}+\dfrac{1}{2^2}+...+\dfrac{1}{2^{2022}}\right)\)
\(=1+\dfrac{1}{2}+...+\dfrac{1}{2^{2021}}-\left(\dfrac{1}{2}+\dfrac{1}{2^2}+...+\dfrac{1}{2^{2022}}\right)\)
\(=1-\dfrac{1}{2^{2022}}\)
b) \(B=\dfrac{20+15+12+17}{60}=\dfrac{4}{5}=1-\dfrac{1}{5}\)
\(A>B\left(Vì\left(\dfrac{1}{2^{2022}}< \dfrac{1}{5}\right)\right)\)
2:
\(B=\left(\dfrac{1}{2^2}-1\right)\left(\dfrac{1}{3^2}-1\right)\cdot...\cdot\left(\dfrac{1}{100^2}-1\right)\)
\(=\left(\dfrac{1}{2}-1\right)\left(\dfrac{1}{2}+1\right)\left(\dfrac{1}{3}-1\right)\left(\dfrac{1}{3}+1\right)\cdot...\cdot\left(\dfrac{1}{100}-1\right)\left(\dfrac{1}{100}+1\right)\)
\(=\left(\dfrac{1}{2}-1\right)\left(\dfrac{1}{3}-1\right)\cdot...\cdot\left(\dfrac{1}{100}-1\right)\left(\dfrac{1}{2}+1\right)\left(\dfrac{1}{3}+1\right)\cdot...\cdot\left(\dfrac{1}{100}+1\right)\)
\(=\dfrac{-1}{2}\cdot\dfrac{-2}{3}\cdot...\cdot\dfrac{-99}{100}\cdot\dfrac{3}{2}\cdot\dfrac{4}{3}\cdot...\cdot\dfrac{101}{100}\)
\(=-\dfrac{1}{100}\cdot\dfrac{101}{2}=\dfrac{-101}{200}< -\dfrac{100}{200}=-\dfrac{1}{2}\)
cả dãy đang trừ mà sao cái cuối là cộng vậy bạn, dãy ko có quy tắc à :v
a, \(A=1+2+2^2+2^3+..........+2^{49}+2^{50}\)
\(\Leftrightarrow2A=2+2^2+..............+2^{50}+2^{51}\)
\(\Leftrightarrow2A-A=\left(2+2^2+.........+2^{51}\right)-\left(1+2+......+2^{50}\right)\)
\(\Leftrightarrow A=2^{51}-1\)
a) \(A=1+2+2^2+2^3+2^4+...+2^{49}+2^{50}\)
\(\Rightarrow2A=2\left(1+2+2^2+2^3+2^4+...+2^{49}+2^{50}\right)\)
\(2A=2+2^2+2^3+2^4+2^5+...+2^{50}+2^{51}\)
\(\Rightarrow2A-A=A=\left(2+2^2+2^3+2^4+2^5+...+2^{50}+2^{51}\right)-\left(1+2+2^2+2^3+2^4+...+2^{49}+2^{50}\right)\)
\(A=2^{51}-1\) vậy \(A=2^{51}-1\)
b) \(B=\dfrac{1}{2}+\left(\dfrac{1}{2}\right)^2+\left(\dfrac{1}{2}\right)^3+\left(\dfrac{1}{2}\right)^4+\left(\dfrac{1}{2}\right)^5+...+\left(\dfrac{1}{2}\right)^{99}+\left(\dfrac{1}{2}\right)^{100}\)
\(B=\dfrac{1}{2}+\dfrac{1}{2^2}+\dfrac{1}{2^3}+\dfrac{1}{2^4}+\dfrac{1}{2^5}+...+\dfrac{1}{2^{99}}+\dfrac{1}{2^{100}}\)
\(\Rightarrow2B=2\left(\dfrac{1}{2}+\dfrac{1}{2^2}+\dfrac{1}{2^3}+\dfrac{1}{2^4}+\dfrac{1}{2^5}+...+\dfrac{1}{2^{99}}+\dfrac{1}{2^{100}}\right)\)
\(2B=\dfrac{2}{2}+\dfrac{2}{2^2}+\dfrac{2}{2^3}+\dfrac{2}{2^4}+\dfrac{2}{2^5}+...+\dfrac{2}{2^{99}}+\dfrac{2}{2^{100}}\)
\(2B=1+\dfrac{1}{2}+\dfrac{1}{2^2}+\dfrac{1}{2^3}+\dfrac{1}{2^4}+...+\dfrac{1}{2^{98}}+\dfrac{1}{2^{99}}\)
\(\Rightarrow2B-B=B=\left(1+\dfrac{1}{2}+\dfrac{1}{2^2}+\dfrac{1}{2^3}+\dfrac{1}{2^4}+...+\dfrac{1}{2^{98}}+\dfrac{1}{2^{99}}\right)-\left(\dfrac{1}{2}+\dfrac{1}{2^2}+\dfrac{1}{2^3}+\dfrac{1}{2^4}+\dfrac{1}{2^5}+\dfrac{1}{2^{99}}+\dfrac{1}{2^{100}}\right)\)
\(B=1-\dfrac{1}{2^{100}}\) vậy \(B=1-\dfrac{1}{2^{100}}\)