Tính tổng sau: \(\dfrac{1}{2^2}+\dfrac{1}{3^2}+...+\dfrac{1}{100^2}< 1\)
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Tính tổng sau: \(S=\dfrac{1}{2+\sqrt{2}}+\dfrac{1}{3\sqrt{2}+2\sqrt{3}}+\dfrac{1}{4\sqrt{3}+3\sqrt{4}}+...+\dfrac{1}{100\sqrt{99}+99\sqrt{100}}\)
Ta có: \(S=\dfrac{1}{2+\sqrt{2}}+\dfrac{1}{3\sqrt{2}+2\sqrt{3}}+\dfrac{1}{4\sqrt{3}+3\sqrt{4}}+...+\dfrac{1}{100\sqrt{99}+99\sqrt{100}}\)
\(=1-\dfrac{1}{\sqrt{2}}+\dfrac{1}{\sqrt{2}}-\dfrac{1}{\sqrt{3}}+...+\dfrac{1}{\sqrt{99}}-\dfrac{1}{10}\)
\(=1-\dfrac{1}{10}=\dfrac{9}{10}\)
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Câu 5: (1 điểm): Tính tổng sau:Q (1-dfrac{1}{2^2}) · (1-dfrac{1}{3^2}) · (1-dfrac{1}{4^2}) · … · (1-dfrac{1}{100^2})
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Câu 5: (1 điểm): Tính tổng sau:
Q = (1-\(\dfrac{1}{2^2}\)) · (1-\(\dfrac{1}{3^2}\)) · (1-\(\dfrac{1}{4^2}\)) · … · (1-\(\dfrac{1}{100^2}\))
\(Q=\left(1-\dfrac{1}{2^2}\right)\left(1-\dfrac{1}{3^2}\right)\cdot...\cdot\left(1-\dfrac{1}{100^2}\right)\)
\(=\left(1-\dfrac{1}{2}\right)\left(1-\dfrac{1}{3}\right)\cdot...\cdot\left(1-\dfrac{1}{100}\right)\left(1+\dfrac{1}{2}\right)\left(1+\dfrac{1}{3}\right)\cdot...\cdot\left(1+\dfrac{1}{100}\right)\)
\(=\dfrac{1}{2}\cdot\dfrac{2}{3}\cdot...\cdot\dfrac{99}{100}\cdot\dfrac{3}{2}\cdot\dfrac{4}{3}\cdot...\cdot\dfrac{101}{100}\)
\(=\dfrac{1}{100}\cdot\dfrac{101}{2}=\dfrac{101}{200}\)
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Tính các tổng sau:
a) A=\(\dfrac{1}{2}+\dfrac{1}{2^2}+\dfrac{1}{2^3}+...+\dfrac{1}{2^{100}}\)
b) B=\(\dfrac{1}{2^2}+\dfrac{1}{3^2}+\dfrac{1}{4^2}+...+\dfrac{1}{102^2}\)
c) C=\(\dfrac{3}{1+2}+\dfrac{3}{1+2+3}+\dfrac{3}{1+2+3+4}+...+\dfrac{3}{1+2+3+...+100}\)
a) \(A=\dfrac{1}{2}+\dfrac{1}{2^2}+............+\dfrac{1}{2^{100}}\)
\(\Leftrightarrow2A=1+\dfrac{1}{2}+\dfrac{1}{2^2}+........+\dfrac{1}{2^{99}}\)
\(\Leftrightarrow2A-A=\left(1+\dfrac{1}{2}+.........+\dfrac{1}{2^{99}}\right)-\left(\dfrac{1}{2}+\dfrac{1}{2^2}+.....+\dfrac{1}{2^{100}}\right)\)
\(\Leftrightarrow A=1-\dfrac{1}{2^{100}}\)
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Mặc dù t cx k biết làm nhưng mà trẩu qá Hằng
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28 tháng 4 2022 lúc 12:32
a)chứng minh rằng :dfrac{1}{3^2}+dfrac{1}{4^2}+dfrac{1}{5^2}+dfrac{1}{6^2}........+dfrac{1}{100^2} dfrac{1}{2}b)tính nhanh tổng S với S dfrac{1}{3.5}+dfrac{1}{5.7}+dfrac{1}{7.9}+......+dfrac{1}{61.63}các cao nhân gải giúp với ạ !!! iem đang cần gấp
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a)chứng minh rằng :\(\dfrac{1}{3^2}\)+\(\dfrac{1}{4^2}\)+\(\dfrac{1}{5^2}\)+\(\dfrac{1}{6^2}\)........+\(\dfrac{1}{100^2}< \dfrac{1}{2}\)
b)tính nhanh tổng S với S= \(\dfrac{1}{3.5}+\dfrac{1}{5.7}+\dfrac{1}{7.9}+......+\dfrac{1}{61.63}\)
các cao nhân gải giúp với ạ !!! iem đang cần gấp
Tính tổng S= \(\dfrac{1}{3}+\dfrac{1}{3^2}+\dfrac{1}{3^3}+\dfrac{1}{3^4}+...+\dfrac{1}{3^{99}}+\dfrac{1}{3^{100}}\)
S = 1/3 + 1/3^2 + 1/3^3 + 1/3^4 + ... + 1/3^99 + 1/3^100
3S = 1 +1/3 +1/3^2 +1/3^3 + ... + 1/3^98 +1/3^99
3S - S = ( 1 + 1/3 + 1/3^2 +1/^3 + ... + 1/3^98 +1/3^99 ) - ( 1/3 + 1/3^2 + 1/3^3 + 1/3^4 +... + 1/3^99 + 1/3^100 )
2S = 1 - 1/3^100
S = (1 - 1/3^100). 1/2
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Bài 1: Tính tổng
\(A=\dfrac{1}{3}+\dfrac{1}{3^2}+\dfrac{1}{3^3}+...+\dfrac{1}{3^{100}}\)
\(A=\dfrac{1}{3}+\dfrac{1}{3^2}+\dfrac{1}{3^3}+...+\dfrac{1}{3^{100}}\\ \Leftrightarrow3A=3\left(+\dfrac{1}{3^2}+\dfrac{1}{3^3}+...+\dfrac{1}{3^{100}}\right)\\ =1+\dfrac{1}{3}+\dfrac{1}{3^2}+...+\dfrac{1}{3^{99}}\)
Lấy 3A - A ta được
\(3A-A=\left(1+\dfrac{1}{3}+\dfrac{1}{3^2}+...+\dfrac{1}{3^{99}}\right)-\left(\dfrac{1}{3}+\dfrac{1}{3^2}+\dfrac{1}{3^3}+...+\dfrac{1}{3^{100}}\right)\\
2A=1-\dfrac{1}{3^{100}}\\
\Leftrightarrow A=\dfrac{1-\dfrac{1}{3^{100}}}{2}\)
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Ta có: \(A=\dfrac{1}{3}+\dfrac{1}{3^2}+\dfrac{1}{3^3}+...+\dfrac{1}{3^{100}}\)
\(\Leftrightarrow3A=1+\dfrac{1}{3}+\dfrac{1}{3^2}+...+\dfrac{1}{3^{99}}\)
\(\Leftrightarrow2\cdot A=1+\dfrac{1}{3}+\dfrac{1}{3^2}+...+\dfrac{1}{3^{99}}-\dfrac{1}{3}-\dfrac{1}{3^2}-...-\dfrac{1}{3^{100}}\)
\(\Leftrightarrow2\cdot A=1-\dfrac{1}{3^{100}}\)
\(\Leftrightarrow2\cdot A=\dfrac{3^{100}-1}{3^{100}}\)
\(\Leftrightarrow A=\dfrac{3^{100}-1}{2\cdot3^{100}}\)
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Thu gọn các tổng sau:
a. A=8.5100.(\(\dfrac{1}{5}+\dfrac{1}{5^2}+...+\dfrac{1}{5^{100}}\)) +1
b. B=\(\dfrac{4}{3}-\dfrac{4}{3^2}+...-\dfrac{4}{3^{100}}\)
a) Đặt \(C=\dfrac{1}{5}+\dfrac{1}{5^2}+...+\dfrac{1}{5^{100}}\)
\(\Rightarrow5C=1+\dfrac{1}{5}+\dfrac{1}{5^2}+...+\dfrac{1}{5^{99}}\)
\(\Rightarrow5C-C=1-\dfrac{1}{5^{100}}\Rightarrow4C=1-\dfrac{1}{5^{100}}\Rightarrow C=\dfrac{1-\dfrac{1}{5^{100}}}{4}\)
\(\Rightarrow A=8.5^{100}.\dfrac{1-\dfrac{1}{5^{100}}}{4}+1=2.\left(5^{100}-1\right)+1=2.5^{100}-2+1=2.5^{100}-1\)
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b)\(B=\dfrac{4}{3}-\dfrac{4}{3^2}+...-\dfrac{4}{3^{100}}\)
\(B=4.\left(\dfrac{1}{3}-\dfrac{1}{3^2}+...-\dfrac{1}{3^{100}}\right)\)
Đặt \(\left(\dfrac{1}{3}-\dfrac{1}{3^2}+...-\dfrac{1}{3^{100}}\right)=D\)
\(\Rightarrow3D=1-\dfrac{1}{3}+...-\dfrac{1}{3^{99}}\)
\(\Rightarrow3D+D=1-\dfrac{1}{3^{100}}\)
\(\Rightarrow D=\dfrac{1-\dfrac{1}{3^{100}}}{4}\)
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Tính giá trị biểu thức sau:
\(D=\left(\dfrac{1}{2^2}-1\right)\left(\dfrac{1}{3^2}-1\right)\left(\dfrac{1}{4^2}-1\right)...\left(\dfrac{1}{100^2}-1\right)\)
Tính các tổng sau:
\(T=\dfrac{1}{1+\sqrt{5}}+\dfrac{1}{\sqrt{5}+\sqrt{9}}+\dfrac{1}{\sqrt{9}+\sqrt{13}+......+\dfrac{1}{\sqrt{2013}+\sqrt{2017}}}\)
\(S=\dfrac{1}{2\sqrt{1}+1\sqrt{2}}+\dfrac{1}{3\sqrt{2}+2\sqrt{3}}+.....+\dfrac{1}{100\sqrt{99}+99\sqrt{100}}\)