CM: \(\dfrac{5}{3.7}+\dfrac{5}{7.11}+\dfrac{5}{11.15}+.....+\dfrac{3}{\left(4n-1\right)\left(4n+3\right)}=\dfrac{5n}{4n+3}\)
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Chứng minh rằng với mọi số tự nhiên khác 0 ta đều có :
a) dfrac{1}{2.5}+dfrac{1}{5.8}+dfrac{1}{8.11}+...+dfrac{1}{left(3n-1right).left(3n+2right)}dfrac{n}{6n+4}
b) dfrac{5}{3.7}+dfrac{5}{7.11}+dfrac{5}{11.15}+...+dfrac{5}{left(4n-1right).left(4n+3right)}dfrac{5n}{4n+3}
giúp mk với
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Chứng minh rằng với mọi số tự nhiên khác 0 ta đều có :
a) \(\dfrac{1}{2.5}+\dfrac{1}{5.8}+\dfrac{1}{8.11}+...+\dfrac{1}{\left(3n-1\right).\left(3n+2\right)}=\dfrac{n}{6n+4}\)
b) \(\dfrac{5}{3.7}+\dfrac{5}{7.11}+\dfrac{5}{11.15}+...+\dfrac{5}{\left(4n-1\right).\left(4n+3\right)}=\dfrac{5n}{4n+3}\)
giúp mk với 
a)
ta có:
\(\left\{{}\begin{matrix}\dfrac{b-a}{b-a}=1..\forall a\ne b\\\dfrac{b-a}{a.b}=\dfrac{1}{a}-\dfrac{1}{b}..\forall a,b\ne0\end{matrix}\right.\)(*)
\(A=\dfrac{1}{2.5}+\dfrac{1}{5.8}+..+\dfrac{1}{\left(3n-1\right)\left(3n+2\right)}\)
\(\left\{{}\begin{matrix}a=3n-1\\b=3n+2\end{matrix}\right.\)\(\Rightarrow b-a=3..\forall n\)
Thay (*) vào dãy A
\(A=\dfrac{1}{3}\left(\dfrac{1}{2}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{8}+\dfrac{1}{8}-\dfrac{1}{11}+\dfrac{1}{11}-....+\dfrac{1}{3n-1}-\dfrac{1}{3n+2}\right)\)
\(A=\dfrac{1}{3}\left(\dfrac{1}{2}-\dfrac{1}{3n+2}\right)=\dfrac{1}{3}\left(\dfrac{3n+2-2}{2.\left(3n+2\right)}\right)=\dfrac{n}{6n+4}=VP\rightarrow dpcm\)
B) tương tự
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Chứng minh \(\frac{5}{3.7}+\frac{5}{7.11}+\frac{5}{11.15}+...+\frac{5}{\left(4n-1\right)\left(4n+3\right)}=\frac{5n}{4n+3}\)
10 tháng 9 2017 lúc 11:58
VTdfrac{1}{1.5}+dfrac{1}{5.9}+...+dfrac{1}{left(4n-3right)left(4n+1right)}
VTdfrac{1}{4}left(1-dfrac{1}{5}+dfrac{1}{5}-dfrac{1}{9}+...+dfrac{1}{4n-3}-dfrac{1}{4n+1}right)
VTdfrac{1}{4}left(1-dfrac{1}{4n+1}right)
VTdfrac{1}{4}left(dfrac{4n}{4n+1}right)
VTdfrac{4n}{16n+4}dfrac{4n}{4left(4n+1right)}dfrac{n}{4n+1}VP
Xảy ra với mọi nin Z^+
@Đặng Thị Cẩm Tú
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\(VT=\dfrac{1}{1.5}+\dfrac{1}{5.9}+...+\dfrac{1}{\left(4n-3\right)\left(4n+1\right)}\)
\(VT=\dfrac{1}{4}\left(1-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{9}+...+\dfrac{1}{4n-3}-\dfrac{1}{4n+1}\right)\)
\(VT=\dfrac{1}{4}\left(1-\dfrac{1}{4n+1}\right)\)
\(VT=\dfrac{1}{4}\left(\dfrac{4n}{4n+1}\right)\)
\(VT=\dfrac{4n}{16n+4}=\dfrac{4n}{4\left(4n+1\right)}=\dfrac{n}{4n+1}=VP\)
Xảy ra với mọi \(n\in Z^+\)
Chứng minh rằng:
a,\(\frac{5}{3.7}+\frac{5}{7.11}+\frac{5}{11.15}+...+\frac{5}{\left(4n-1\right).\left(4n+3\right)}=\frac{5n}{3.\left(4n+3\right)}\)
b,\(\frac{1}{1+2+3}+\frac{1}{1+2+3+4}+...+\frac{1}{1+2+3+...+100}< \frac{1}{4}\)
\(lim\dfrac{4n^5+n^3-5n+3}{\left(2n^2-3\right)\left(2n+1\right)^3}\)
\(lim\dfrac{4n^5+n^3-5n+3}{\left(2n^2-3\right)\left(2n+1\right)^3}\)
Lời giải:
\(\lim \frac{4n^5+n^3-5n+3}{(2n^2-3)(2n+1)^3}=\lim \frac{4+\frac{1}{n^2}-\frac{5}{n^4}+\frac{3}{n^5}}{(2-\frac{3}{n^2})(2+\frac{1}{n})^3}=\frac{4}{2.2^3}=\frac{1}{4}\)
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\(CMR:\frac{5}{3.7}+\frac{5}{7.11}+...+\frac{5}{\left(4n-1\right)\left(4n+3\right)}=\frac{5n}{4n+3}\)
Tìm các giới hạn sau:
\(a,\dfrac{4n^5-3n^2}{\left(3n^2-2\right)\left(1-4n^3\right)}\)
\(b,\dfrac{\left(n^2+1\right)\left(n-10\right)^2}{\left(n+1\right)\left(3n-3\right)^3}\)
\(b,lim\dfrac{\left(n^2+1\right)\left(n-10\right)^2}{\left(n+1\right)\left(3n-3\right)^3}\)
\(=lim\dfrac{\left(1+\dfrac{1}{n^2}\right)\left(\dfrac{1}{n}-\dfrac{10}{n^2}\right)^2}{\left(1+\dfrac{1}{n}\right)\left(\dfrac{3}{n^2}-\dfrac{3}{n^3}\right)}=0\)
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\(a,lim\dfrac{4n^5-3n^2}{\left(3n^2-2\right)\left(1-4n^3\right)}\)
\(=lim\dfrac{4-\dfrac{3}{n^3}}{\left(3-\dfrac{2}{n^2}\right)\left(\dfrac{1}{n^3}-4\right)}\)
\(=\dfrac{4-0}{\left(3-0\right)\left(0-4\right)}=\dfrac{4}{-12}=-\dfrac{1}{3}\)
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\(\lim\dfrac{\left(n^2+1\right)\left(n-10\right)^2}{\left(n+1\right)\left(3n-3\right)^3}=\lim\dfrac{\left(1+\dfrac{1}{n^2}\right)\left(1-\dfrac{10}{n}\right)^2}{\left(1+\dfrac{1}{n}\right)\left(3-\dfrac{3}{n}\right)^3}=\dfrac{1.1^2}{1.3}=\dfrac{1}{3}\)
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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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