tính: \(\sqrt{1+\dfrac{1}{2^2}+\dfrac{1}{3^2}}+\sqrt{1+\dfrac{1}{2^2}+\dfrac{1}{3^2}}+....+\sqrt{1+\dfrac{1}{\left(n-1\right)^2}+\dfrac{1}{n^2}}\)
Những câu hỏi liên quan
1) Chứng minh rằng: 1+dfrac{1}{2sqrt{2}}+dfrac{1}{3sqrt{3}}+...+dfrac{1}{nsqrt{n}} 2sqrt{2}left(nin Nright)
2) Chứng minh rằng: dfrac{2}{3}+sqrt{n+1} 1+sqrt{2}+sqrt{3}+...+sqrt{n} dfrac{2}{3}left(n+1right)sqrt{n}
3) 2sqrt{n}-3 dfrac{1}{sqrt{2}}+dfrac{1}{sqrt{3}}+...+dfrac{1}{sqrt{n}} 2sqrt{n}-2
4) dfrac{sqrt{2}-sqrt{1}}{2+1}+dfrac{sqrt{3}-sqrt{2}}{3+2}+...+dfrac{sqrt{n+1}-sqrt{n}}{n+1+n} dfrac{1}{2}left(1-dfrac{1}{sqrt{n+1}}right)
Đọc tiếp
1) Chứng minh rằng: \(1+\dfrac{1}{2\sqrt{2}}+\dfrac{1}{3\sqrt{3}}+...+\dfrac{1}{n\sqrt{n}}< 2\sqrt{2}\left(n\in N\right)\)
2) Chứng minh rằng: \(\dfrac{2}{3}+\sqrt{n+1}< 1+\sqrt{2}+\sqrt{3}+...+\sqrt{n}< \dfrac{2}{3}\left(n+1\right)\sqrt{n}\)
3) \(2\sqrt{n}-3< \dfrac{1}{\sqrt{2}}+\dfrac{1}{\sqrt{3}}+...+\dfrac{1}{\sqrt{n}}< 2\sqrt{n}-2\)
4) \(\dfrac{\sqrt{2}-\sqrt{1}}{2+1}+\dfrac{\sqrt{3}-\sqrt{2}}{3+2}+...+\dfrac{\sqrt{n+1}-\sqrt{n}}{n+1+n}< \dfrac{1}{2}\left(1-\dfrac{1}{\sqrt{n+1}}\right)\)
CMR:
\(\dfrac{1}{2\sqrt{1}+1\sqrt{2}}+\dfrac{1}{3\sqrt{2}+2\sqrt{3}}+\dfrac{1}{4\sqrt{3}+3\sqrt{3}}+....+\dfrac{1}{\left(n+1\right)\left(\sqrt{n}+n\sqrt{n+1}\right)}< 1\)
Tính giới hạn của dãy số:
\(u_n=\dfrac{1}{2\sqrt{1}+\sqrt{2}}+\dfrac{1}{3\sqrt{2}+2\sqrt{3}}+...+\dfrac{1}{\left(n+1\right)\sqrt{n}+n\sqrt{n+1}}\)
Một câu thôi: Liên hợp
\(\dfrac{1}{2\sqrt{1}+\sqrt{2}}=\dfrac{2.1-\sqrt{2}}{2^2-2}=\dfrac{2-\sqrt{2}}{2}=1-\dfrac{1}{\sqrt{2}}\)
\(\dfrac{1}{3\sqrt{2}+2\sqrt{3}}=\dfrac{3\sqrt{2}-2\sqrt{3}}{9.2-4.3}=\dfrac{3\sqrt{2}-2\sqrt{3}}{6}=\dfrac{1}{\sqrt{2}}-\dfrac{1}{\sqrt{3}}\)
\(\Rightarrow\dfrac{1}{\left(n+1\right)\sqrt{n}+n\sqrt{n+1}}=\dfrac{1}{\sqrt{n}}-\dfrac{1}{\sqrt{n+1}}\)
Nên chứng minh bằng quy nạp mạnh cho chặt chẽ, giờ tui buồn ngủ quá nên bạn tự chứng minh nha :(
\(\Rightarrow u_n=1-\dfrac{1}{\sqrt{n+1}}=\dfrac{\sqrt{n+1}-1}{\sqrt{n+1}}\Rightarrow\lim\limits\left(u_n\right)=\lim\limits\dfrac{\sqrt{\dfrac{n}{n}+\dfrac{1}{n}}-\dfrac{1}{\sqrt{n}}}{\sqrt{\dfrac{n}{n}+\dfrac{1}{n}}}=1\)
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Cho n ϵ N*. Chứng minh:
a) \(1+\dfrac{1}{2^2}+\dfrac{1}{3^2}+...+\dfrac{1}{\left(n-1\right)^2}+\dfrac{1}{n^2}< 2\)
b) \(1+\dfrac{1}{\sqrt{2}}+\dfrac{1}{\sqrt{3}}+...+\dfrac{1}{\sqrt{n}}>2\left(\sqrt{n+1}-1\right)\)
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Áp dụng : \(\dfrac{1}{\sqrt{n}}>2\left(\sqrt{n+1}-\sqrt{n}\right)\)
\(\dfrac{1}{\sqrt{n}}+\dfrac{1}{\sqrt{n-1}}+...+\dfrac{1}{\sqrt{3}}+\dfrac{1}{\sqrt{2}}+1>2\left(\sqrt{n+1}-\sqrt{n}\right)+2\left(\sqrt{n}-\sqrt{n-1}\right)+...+2\left(\sqrt{4}-\sqrt{3}\right)+2\left(\sqrt{3}-\sqrt{2}\right)+2\left(\sqrt{2}-1\right).\)
\(=2\left(\sqrt{n+1}-1\right).\)
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Tính các giới hạn
a) \(lim\dfrac{1+\dfrac{1}{3}+\left(\dfrac{1}{3}\right)^2+...+\left(\dfrac{1}{3}\right)^n}{1+\dfrac{1}{2}+\left(\dfrac{1}{2}\right)^2+...+\left(\dfrac{1}{2}\right)^n}\)
\(lim\left(n^3+n\sqrt{n}-5\right)\)
Giúp mình với ạ
a/ \(\lim\limits\dfrac{1+\dfrac{1}{3}+\left(\dfrac{1}{3}\right)^2+...+\left(\dfrac{1}{3}\right)^n}{1+\dfrac{1}{2}+\left(\dfrac{1}{2}\right)^2+...+\left(\dfrac{1}{2}\right)^n}=\lim\limits\dfrac{\dfrac{\left(\dfrac{1}{3}\right)^{n+1}-1}{\dfrac{1}{3}-1}}{\dfrac{\left(\dfrac{1}{2}\right)^{n+1}-1}{\dfrac{1}{2}-1}}=\dfrac{\dfrac{3}{2}}{\dfrac{1}{2}}=3\)
b/ \(\lim\limits\left(n^3+n\sqrt{n}-5\right)=+\infty-5=+\infty\)
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23 tháng 9 2021 lúc 20:50
Tính \(S=\sqrt{1+\dfrac{8.1^2-1}{1^2.3^2}}+\sqrt{1+\dfrac{8.2^2-1}{3^2.5^2}}+...+\sqrt{1+\dfrac{8.n^2-1}{\left(2n-1\right)^2.\left(2n+1\right)^2}}\)
Với\(n\in N\)
Tính các giới hạn sau
1,Lim\(\left(\dfrac{2n^3}{2n^2+3}+\dfrac{1-5n^2}{5n+1}\right)\)
2,a,Lim\(\left(\sqrt{n^2+n}-\sqrt{n^2+2}\right)\)
b,Lim\(\dfrac{\sqrt{n^4+3n-2}}{2n^2-n+3}\)
c,Lim\(\dfrac{\sqrt{n^2-4n}-\sqrt{4n^2+1}}{\sqrt{3n^2+1}-n}\)
\(a=\lim\left(\dfrac{2n^3\left(5n+1\right)+\left(2n^2+3\right)\left(1-5n^2\right)}{\left(2n^2+3\right)\left(5n+1\right)}\right)\)
\(=\lim\left(\dfrac{2n^3-13n^2+3}{\left(2n^2+3\right)\left(5n+1\right)}\right)=\lim\dfrac{2-\dfrac{13}{n}+\dfrac{3}{n^3}}{\left(2+\dfrac{3}{n^2}\right)\left(5+\dfrac{1}{n}\right)}=\dfrac{2}{2.5}=\dfrac{1}{5}\)
\(b=\lim\left(\dfrac{n-2}{\sqrt{n^2+n}+\sqrt{n^2+2}}\right)=\lim\dfrac{1-\dfrac{2}{n}}{\sqrt{1+\dfrac{1}{n}}+\sqrt{1+\dfrac{2}{n}}}=\dfrac{1}{2}\)
\(c=\lim\dfrac{\sqrt{1+\dfrac{3}{n^3}-\dfrac{2}{n^4}}}{2-\dfrac{2}{n}+\dfrac{3}{n^2}}=\dfrac{1}{2}\)
\(d=\lim\dfrac{\sqrt{1-\dfrac{4}{n}}-\sqrt{4+\dfrac{1}{n^2}}}{\sqrt{3+\dfrac{1}{n^2}}-1}=\dfrac{1-2}{\sqrt{3}-1}=-\dfrac{1+\sqrt{3}}{2}\)
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19 tháng 9 2021 lúc 21:42
Cdfrac{1}{1+sqrt{2}}+dfrac{1}{sqrt{2}+sqrt{3}}+dfrac{1}{sqrt{3}+sqrt{4}}+....dfrac{1}{sqrt{99}+sqrt{100}}Cdfrac{1left(1-sqrt{2}right)}{left(1+sqrt{2}right)left(1-sqrt{2}right)}+dfrac{1left(sqrt{2}-sqrt{3}right)}{left(sqrt{2}-sqrt{3}right)left(sqrt{2}+sqrt{3}right)}+dfrac{1left(sqrt{3}-sqrt{4}right)}{left(sqrt{3}-sqrt{4}right)left(sqrt{3}+sqrt{4}right)}+........dfrac{1left(sqrt{99}-sqrt{100}right)}{left(sqrt{99}-sqrt{100}right)left(sqrt{99}+sqrt{100}right)}Cdfrac{1-sqrt{2}}{1-2}+dfrac{sqrt{2}-sqr...
Đọc tiếp
\(C=\dfrac{1}{1+\sqrt{2}}+\dfrac{1}{\sqrt{2}+\sqrt{3}}+\dfrac{1}{\sqrt{3}+\sqrt{4}}+....\dfrac{1}{\sqrt{99}+\sqrt{100}}\)
\(C=\dfrac{1\left(1-\sqrt{2}\right)}{\left(1+\sqrt{2}\right)\left(1-\sqrt{2}\right)}+\dfrac{1\left(\sqrt{2}-\sqrt{3}\right)}{\left(\sqrt{2}-\sqrt{3}\right)\left(\sqrt{2}+\sqrt{3}\right)}+\dfrac{1\left(\sqrt{3}-\sqrt{4}\right)}{\left(\sqrt{3}-\sqrt{4}\right)\left(\sqrt{3}+\sqrt{4}\right)}+........\dfrac{1\left(\sqrt{99}-\sqrt{100}\right)}{\left(\sqrt{99}-\sqrt{100}\right)\left(\sqrt{99}+\sqrt{100}\right)}\)
\(C=\dfrac{1-\sqrt{2}}{1-2}+\dfrac{\sqrt{2}-\sqrt{3}}{2-3}+\dfrac{\sqrt{3}-\sqrt{4}}{3-4}+.....+\dfrac{\sqrt{99}-\sqrt{100}}{99-100}\)
\(C=\dfrac{1-\sqrt{2}}{-1}+\dfrac{\sqrt{2}-\sqrt{3}}{-1}+\dfrac{\sqrt{3}-\sqrt{4}}{-1}+......+\dfrac{\sqrt{99}-\sqrt{100}}{-1}\)
\(C=-\left(1-\sqrt{2}\right)-\left(\sqrt{2}+\sqrt{3}\right)-\left(\sqrt{3}-\sqrt{4}\right)-......-\left(\sqrt{99}-\sqrt{100}\right)\)
\(C=-1+\sqrt{2}-\sqrt{2}+\sqrt{3}-\sqrt{3}+\sqrt{4}-......-\sqrt{99}+\sqrt{100}\)
\(C=-1+\sqrt{100}\)
\(C=10-1=9\)
trong bai :
cho a dfrac{1}{2sqrt{1}+1sqrt{2}}+dfrac{1}{3sqrt{2}+2sqrt{3}}+dfrac{1}{4sqrt{3}+3sqrt{4}}+....+dfrac{1}{left(n+1right)sqrt{n}+nsqrt{n+1}} 1
co phan huong dan : dfrac{1}{left(n+1right)sqrt{n}+nsqrt{n+1}}dfrac{1}{sqrt{n}.sqrt{n+1}left(sqrt{n+1}+sqrt{n}right)}dfrac{sqrt{n+1}-sqrt{n}}{sqrt{n}.sqrt{n+1}}dfrac{1}{sqrt{n}}-dfrac{1}{sqrt{n+1}}
cho minh hoi buoc : dfrac{sqrt{n+1}-sqrt{n}}{sqrt{n}.sqrt{n+1}} tu dau ra .( giai thich chi tiet)
Đọc tiếp
trong bai :
cho a= \(\dfrac{1}{2\sqrt{1}+1\sqrt{2}}+\dfrac{1}{3\sqrt{2}+2\sqrt{3}}+\dfrac{1}{4\sqrt{3}+3\sqrt{4}}+....+\dfrac{1}{\left(n+1\right)\sqrt{n}+n\sqrt{n+1}}< 1\)
co phan huong dan : \(\dfrac{1}{\left(n+1\right)\sqrt{n}+n\sqrt{n+1}}=\dfrac{1}{\sqrt{n}.\sqrt{n+1}\left(\sqrt{n+1}+\sqrt{n}\right)}=\dfrac{\sqrt{n+1}-\sqrt{n}}{\sqrt{n}.\sqrt{n+1}}=\dfrac{1}{\sqrt{n}}-\dfrac{1}{\sqrt{n+1}}\)
cho minh hoi buoc : \(\dfrac{\sqrt{n+1}-\sqrt{n}}{\sqrt{n}.\sqrt{n+1}}\) tu dau ra .( giai thich chi tiet)
\(\dfrac{1}{\sqrt{n}.\sqrt{n+1}.\left(\sqrt{n+1}+\sqrt{n}\right)}=\dfrac{1}{\sqrt{n}.\sqrt{n+1}}.\dfrac{1}{\sqrt{n+1}+\sqrt{n}}=\dfrac{1}{\sqrt{n}.\sqrt{n+1}}.\dfrac{\sqrt{n+1}-\sqrt{n}}{n+1-n}=\dfrac{1}{\sqrt{n}.\sqrt{n+1}}.\left(\sqrt{n+1}-\sqrt{n}\right)=\dfrac{\sqrt{n+1}-\sqrt{n}}{\sqrt{n}.\sqrt{n+1}}\)
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