Tìm n?
\(3C^o_{2n}-\dfrac{1}{2}C^1_{2n}-\dfrac{1}{4}C^3_{2n}+...+\dfrac{3}{2n+1}C^{2n}_{2n}=\dfrac{10923}{5}\)
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AH
Akai Haruma
Giáo viên
20 tháng 1 2018
Lời giải:
Theo nhị thức New-ton:
\((x+1)^{2n}=C^{0}_{2n}+C^{1}_{2n}x+C^2_{2n}x^2+...+C^{2n}_{2n}x^{2n}\)
\((x-1)^n=C^0_{2n}-C^1_{2n}x+C^2_{2n}x^2-.....-C^{2n-1}_{2n}x^{2n-1}+C^{2n}_{2n}x^{2n}\)
Trừ theo vế ta có:
\(\frac{(x+1)^{2n}-(x-1)^{2n}}{2}=C^1_{2n}x+C^3_{2n}x^3+...+C^{2n-1}_{2n}x^{2n-1}\)
\(\Rightarrow \int ^{1}_{0}\frac{(x+1)^{2n}-(x-1)^{2n}}{2}dx=\int ^{1}_{0}(C^1_{2n}x+C^3_{2n}x^3+...+C^{2n-1}_{2n}x^{2n-1})dx\)
Xét vế trái:
\(\text{VT}=\frac{1}{2}\int ^{1}_{0}(x+1)^{2n}d(x+1)-\frac{1}{2}\int ^{1}_{0}(x-1)^{2n}d(x-1)\)
\(=\left.\begin{matrix} 1\\ 0\end{matrix}\right|\frac{1}{2}\left ( \frac{(x+1)^{2n+1}-(x-1)^{2n+1}}{2n+1} \right )=\frac{2^{2n}-1}{2n+1}\)
Xét vế phải:
\(\text{VP}=\left.\begin{matrix} 1\\ 0\end{matrix}\right|\left ( \frac{C^{1}_{2n}x^2}{2}+\frac{C^{3}_{2n}x^4}{4}+....+\frac{C^{2n-1}_{2n}x^{2n}}{2n} \right )=\frac{1}{2}C^{1}_{2n}+\frac{1}{4}C^3_{2n}+...+\frac{1}{2n}C^{2n-1}_{2n}\)
Vậy \(A=\frac{2^{2n}-1}{2n+1}\)
MH
12 tháng 2 2022
\(a,lim\dfrac{2n^2+1}{3n^3-3n+3}\)
\(=lim\dfrac{\dfrac{2}{n}+\dfrac{1}{n^3}}{3-\dfrac{3}{n^2}+\dfrac{3}{n^3}}=0\)
NV
Nguyễn Việt Lâm
Giáo viên
12 tháng 2 2022
\(\lim\dfrac{-3n^3+1}{2n+5}=\lim\dfrac{-3n^2+\dfrac{1}{n}}{2+\dfrac{5}{n}}=\dfrac{-\infty}{2}=-\infty\)
\(\lim\dfrac{n^3-2n+1}{-3n-4}=\lim\dfrac{n^2-2+\dfrac{1}{n}}{-3-\dfrac{4}{n}}=\dfrac{+\infty}{-3}=-\infty\)
NV
Nguyễn Việt Lâm
Giáo viên
6 tháng 2 2021
\(a=\lim n\left(\sqrt[3]{-1+\dfrac{2}{n}-\dfrac{5}{n^3}}\right)=+\infty.\left(-1\right)=-\infty\)
\(b=\lim\left(\sqrt{n+1}+\sqrt{n}\right)=+\infty\)
\(c=\lim n\left(\dfrac{1}{n^2+n}-1\right)=+\infty.\left(-1\right)=-\infty\)
\(d=\lim\left(\dfrac{2n^2-1-2n\left(n+1\right)}{n+1}\right)=\lim\left(\dfrac{-1-2n}{n+1}\right)=-2\)
\(e=\lim\dfrac{2n^2+n-3+\dfrac{1}{n}}{\dfrac{2}{n}-3}=\dfrac{+\infty}{-3}=-\infty\)
AH
Akai Haruma
Giáo viên
21 tháng 4 2018
Lời giải:
Ta có:
\(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=\frac{1}{a+b+c}\)
\(\Leftrightarrow \frac{a+b}{ab}=\frac{1}{a+b+c}-\frac{1}{c}=\frac{-(a+b)}{c(a+b+c)}\)
\(\Leftrightarrow (a+b)\left(\frac{1}{ab}+\frac{1}{c(a+b+c)}\right)=0\)
\(\Leftrightarrow (a+b).\frac{ab+c(a+b+c)}{abc(a+b+c)}=0\)
\(\Leftrightarrow (a+b).\frac{(c+a)(c+b)}{abc(a+b+c)}=0\)
\(\Leftrightarrow (a+b)(b+c)(c+a)=0\)
Ta sẽ cm \(\frac{1}{a^{2n+1}}+\frac{1}{b^{2n+1}}+\frac{1}{c^{2n+1}}=\frac{1}{a^{2n+1}+b^{2n+1}+c^{2n+1}}(*)\)
Thật vậy: \((*)\Leftrightarrow \frac{a^{2n+1}+b^{2n+1}}{(ab)^{2n+1}}=\frac{1}{a^{2n+1}+b^{2n+1}+c^{2n+1}}-\frac{1}{c^{2n+1}}\)
\(\Leftrightarrow \frac{a^{2n+1}+b^{2n+1}}{(ab)^{2n+1}}=\frac{-(a^{2n+1}+b^{2n+1})}{c^{2n+1}(a^{2n+1}+b^{2n+1}+c^{2n+1})}\)
\(\Leftrightarrow (a^{2n+1}+b^{2n+1})\left(\frac{1}{(ab)^{2n+1)}}+\frac{1}{c^{2n+1}(a^{2n+1}+b^{2n+1}+c^{2n+1})}\right)=0\)
\(\Leftrightarrow (a^{2n+1}+b^{2n+1}).\frac{c^{2n+1}(a^{2n+1}+b^{2n+1}+c^{2n+1})+(ab)^{2n+1}}{(abc)^{2n+1}(a^{2n+1}+b^{2n+1}+c^{2n+1})}=0\)
\(\Leftrightarrow \frac{(a^{2n+1}+b^{2n+1})(c^{2n+1}+b^{2n+1})(c^{2n+1}+a^{2n+1})}{abc^{2n+1}(a^{2n+1}+b^{2n+1}+c^{2n+1})}=0\)
Thấy rằng
\((a^{2n+1}+b^{2n+1})(b^{2n+1}+c^{2n+1})(c^{2n+1}+a^{2n+1})=(a+b).X.(b+c).Y.(c+a).Z\)
\(=0\) (do \((a+b)(b+c)(c+a)=0\) )
Do đó đẳng thức $(*)$ cần chứng minh đúng.
-------------------
Ta tiếp tục chứng minh \(\frac{1}{a^{2n+1}+b^{2n+1}+c^{2n+1}}=\frac{1}{(a+b+c)^{2n+1}}(**)\)
\(\Leftrightarrow a^{2n+1}+b^{2n+1}+c^{2n+1}=(a+b+c)^{2n+1}\)
Thật vậy:
\((a+b)(b+c)(c+a)=0\)\(\Rightarrow \left[\begin{matrix} a+b=0\\ b+c=0\\ c+a=0\end{matrix}\right.\)
Không mất tổng quát giả sử \(a+b=0\)
\(\Rightarrow \left\{\begin{matrix} a^{2n+1}+b^{2n+1}+c^{2n+1}=(-b)^{2n+1}+b^{2n+1}+c^{2n+1}=c^{2n+1}\\ (a+b+c)^{2n+1}=(0+c)^{2n+1}=c^{2n+1}\end{matrix}\right.\)
\(\Rightarrow a^{2n+1}+b^{2n+1}+c^{2n+1}=(a+b+c)^{2n+1}\)
Do đó $(**)$ đúng
Từ $(*)$ và $(**)$ ta có đpcm.
HN
23 tháng 4 2018
Ta có:
\(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}=\dfrac{1}{a+b+c}\)
\(\Leftrightarrow\left(a+b\right)\left(b+c\right)\left(c+a\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}a=-b\\b=-c\\c=-a\end{matrix}\right.\)
Xét \(a=-b\) thì ta có
\(\left\{{}\begin{matrix}\dfrac{1}{a^{2n+1}}+\dfrac{1}{b^{2n+1}}+\dfrac{1}{c^{2n+1}}=\dfrac{1}{c^{2n+1}}\\\dfrac{1}{a^{2n+1}+b^{2n+1}+c^{2n+1}}=\dfrac{1}{c^{2n+1}}\\\dfrac{1}{\left(a+b+c\right)^{2n+1}}=\dfrac{1}{c^{2n+1}}\end{matrix}\right.\)
\(\Rightarrow\dfrac{1}{a^{2n+1}}+\dfrac{1}{b^{2n+1}}+\dfrac{1}{c^{2n+1}}=\dfrac{1}{a^{2n+1}+b^{2n+1}+c^{2n+1}}=\dfrac{1}{\left(a+b+c\right)^{2n+1}}\)
Tương tự cho 2 bộ số còn lại ta được ĐPCM.
MH
14 tháng 9 2021
c)\(7^{2n}+7^{2n+2}=2450\)
⇒\(7^{2n}+7^{2n}.7^2=2450\)
⇒\(7^{2n}.50=2450\)
⇒\(7^{2n}=49\)\(=7^2\)
⇒2n=2
⇒n=1
10 tháng 2 2021
a,\(lim\dfrac{n^2-2n}{5n+3n^2}=lim\dfrac{1-\dfrac{2}{n}}{\dfrac{5}{n}+3}=\dfrac{1}{3}\)
b,\(lim\dfrac{n^2-2}{5n+3n^2}=lim\dfrac{1-\dfrac{2}{n^2}}{\dfrac{5}{n}+3}=\dfrac{1}{3}\)
c,\(lim\dfrac{1-2n}{5n+3n^2}=lim\dfrac{1-2n}{n\left(5+3n\right)}=lim\dfrac{\dfrac{1}{n}-2}{1\left(\dfrac{5}{n}+3\right)}=-\dfrac{2}{3}\)
d,\(lim\dfrac{1-2n^2}{5n+5}=lim\dfrac{\left(1-n\sqrt{2}\right)\left(1+n\sqrt{2}\right)}{5n+5}=lim\dfrac{\left(\dfrac{1}{n}-\sqrt{2}\right)\left(\dfrac{1}{n}+\sqrt{2}\right)}{5+\dfrac{5}{n}}=\dfrac{-2}{5}\)
what are you doing
s bi loi nhi?
tim n?
3C\(^0\)\(_{2n}\) \(-\) \(\dfrac{1}{2}\)C\(^1\)\(_{2n}\) \(-\) \(\dfrac{1}{4}\)C\(^3\)\(_{2n}\) +...+ \(\dfrac{3}{2n+1}\)C\(^{2n}\)\(_{2n}\) \(=\) \(\dfrac{10923}{5}\)