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Tính tổng :
a) 12+322+523+....+2n−12n12+322+523+....+2n−12n
b) 12−22+32−42+....+(−1)n−1.n\(^2\)
Giải
a) HD: Đặt tổng là S\(_n\) và tính 2S\(_n\)
ĐS : S\(_n\)=3−\(\frac{2n+3}{2^n}\)
b) HD: n\(^2\)- (n+1)\(^2\)= -2n-1
Ta có: 1\(^2\)-2\(^2\)= -3; 3\(^2\) - 4\(^2\)= -7;....
Ta có: u\(_1\)= -3, d= -4 và tính S\(_n\) trong từng trường hợp n chẵn, lẻ.
Sn=3−2n+32nb) HD : b) HD : n2−(n+1)2=−2n−1n2−(n+1)2=−2n−1 Ta có 12−22=−3;32−42=−7;...12−22=−3;32−42=−7;... b) HD :
\(A=lim\frac{\sqrt{n+2}+\sqrt{n+1}}{1}=lim\left[n\left(\sqrt{1+\frac{2}{n}}+\sqrt{1+\frac{1}{n}}\right)\right]=+\infty.2=+\infty\)
\(B=lim\frac{8^3.64^n-9.27^n}{4^4.64^n+5^3.25^n}=\frac{8^3-9.\left(\frac{27}{64}\right)^n}{4^4+5^3\left(\frac{25}{64}\right)^n}=\frac{8^3}{4^4}=2\)
\(1;-\frac{1}{2};\frac{1}{4}...\) là dãy cấp số nhân lùi vô hạn có \(u_1=1\) và \(q=-\frac{1}{2}\)
Do \(\left|q\right|< 1\) nên theo công thức tổng cấp số nhân:
\(S_n=\frac{u_1}{1-q}=\frac{1}{1+\frac{1}{2}}=\frac{2}{3}\)
Bài 1. Ta có:
\(\begin{array}{l} S = \sum\limits_{k = 1}^n {{x^{2k}}} + \sum\limits_{k = 1}^n {\dfrac{1}{{{x^{2k}}}} + 2n} \\ = {x^2}\dfrac{{1 - {x^{2n}}}}{{1 - {x^2}}} + \dfrac{1}{{{x^2}}}.\dfrac{{1 - \dfrac{1}{{{x^{2n}}}}}}{{1 - \dfrac{1}{{{x^2}}}}} + 2n\\ = \dfrac{{\left( {1 - {x^{2n}}} \right)\left( {{x^{2n + 2}} - 1} \right)}}{{\left( {1 - {x^2}} \right){x^{2n}}}} + 2n \end{array}\)
Bài 2.
Ta có:
\(\begin{array}{l} T = \dfrac{1}{2} + \dfrac{3}{{{2^2}}} + \dfrac{5}{{{2^3}}} + ... + \dfrac{{2n - 1}}{{{2^n}}}\left( 1 \right)\\ \dfrac{1}{2}T = \dfrac{1}{{{2^2}}} + \dfrac{3}{{{2^3}}} + \dfrac{5}{{{2^4}}} + ... + \dfrac{{2n - 3}}{{{2^n}}} + \dfrac{{2n - 1}}{{{2^{n + 1}}}}\left( 2 \right) \end{array}\)
\((1)-(2)\)\(\Rightarrow \dfrac{1}{2}T = \dfrac{1}{2} + \dfrac{2}{{{2^2}}} + \dfrac{2}{{{2^3}}} + ... + \dfrac{2}{{{2^n}}} - \dfrac{{2n - 1}}{{{2^{n + 1}}}}\)
\(\begin{array}{l} \Rightarrow T = 2\left[ {\dfrac{1}{2} + \dfrac{1}{2}\dfrac{{1 - {{\left( {\dfrac{1}{2}} \right)}^{n - 1}}}}{{1 - \dfrac{1}{2}}} - \dfrac{{2n - 1}}{{{2^{n + 1}}}}} \right]\\ = 1 + \dfrac{{{2^{n - 1}} - 1}}{{{2^{n - 2}}}} - \dfrac{{2n - 1}}{{{2^n}}} \end{array}\)
\(S=x^2+\frac{1}{x^2}+2+x^4+\frac{1}{x^4}+2+...+x^{2n}+\frac{1}{x^{2n}}+2\)
\(=\left(x^2+x^4+...+x^{2n}\right)+\left(\frac{1}{x^2}+\frac{1}{x^4}+...+\frac{1}{x^{2n}}\right)+2n\)
\(=x^2.\frac{\left(x^2\right)^{n-1}-1}{x^2-1}+\frac{1}{x^2}.\frac{\left(\frac{1}{x^2}\right)^{n-1}-1}{\frac{1}{x^2}-1}+2n\)
\(=\frac{x^{2n}-x^2}{x^2-1}+\frac{x^{2-2n}-1}{1-x^2}+2n\)
\(T=\frac{1}{2}+\frac{3}{2^2}+\frac{5}{2^3}+...+\frac{2n-3}{2^{n-1}}+\frac{2n-1}{2^n}\)
\(\Rightarrow2T=1+\frac{3}{2}+\frac{5}{2^2}+...+\frac{2n-1}{2^{n-1}}\)
\(\Rightarrow T=1+\frac{2}{2}+\frac{2}{2^2}+\frac{2}{2^3}+...+\frac{2}{2^{n-1}}-\frac{2n-1}{2^n}\)
\(T=1+1+\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^{n-2}}-\frac{2n-1}{2^n}\)
\(T=1+1.\frac{\left(\frac{1}{2}\right)^{n-2}-1}{\frac{1}{2}-1}-\frac{2n-1}{2^n}=3-\frac{1}{2^{n-1}}-\frac{2n-1}{2^n}=3-\frac{1}{2^n}-\frac{n}{2^{n-1}}\)
a/ \(lim\left(\sqrt[3]{n-n^3}+n+\sqrt{n^2+3n}-n\right)\)
\(=lim\left(\frac{n}{\sqrt[3]{\left(n-n^3\right)^2}-n\sqrt[3]{\left(n-n^3\right)}+n^2}+\frac{3n}{\sqrt{n^2+3n}+n}\right)\)
\(=lim\left(\frac{1}{\sqrt[3]{n^3+2n+\frac{1}{n}}+\sqrt[3]{n^3-n}+n}+\frac{3}{\sqrt{1+\frac{3}{n}}+1}\right)=0+\frac{3}{1+1}=\frac{3}{2}\)
b/ \(lim\left(\frac{-2\sqrt{n}-4}{\sqrt{n-2\sqrt{n}}+\sqrt{n+4}}\right)=lim\left(\frac{-2-\frac{4}{\sqrt{n}}}{\sqrt{1-\frac{2}{\sqrt{n}}}+\sqrt{1+\frac{4}{n}}}\right)=-\frac{2}{1+1}=-1\)
c/ \(lim\left(\frac{3n^2}{\sqrt[3]{n^6+6n^5+9n^4}+\sqrt[3]{n^6+3n^5}+n^2}\right)=lim\left(\frac{3}{\sqrt[3]{1+\frac{6}{n}+\frac{9}{n^2}}+\sqrt[3]{1+\frac{3}{n}}+1}\right)=\frac{3}{3}=1\)
d/ \(lim\left(\sqrt[3]{n^3+6n}-n+n-\sqrt{n^2-4n}\right)=lim\left(\frac{6n}{\sqrt[3]{n^6+12n^4+36n^2}+\sqrt[3]{n^6+6n^4}+n^2}+\frac{4n}{n+\sqrt{n^2-4n}}\right)\)
\(=lim\left(\frac{6}{\sqrt[3]{n^3+12n+\frac{36}{n}}+\sqrt[3]{n^3+6n}+n}+\frac{4}{1+\sqrt{1-\frac{4}{n}}}\right)=0+\frac{4}{1+1}=2\)
e/ \(lim\left(\frac{-3.3^n+4.4^n}{5.3^n+\frac{3}{2}.4^n}\right)=lim\left(\frac{-3\left(\frac{3}{4}\right)^n+4}{5.\left(\frac{3}{4}\right)^n+\frac{3}{2}}\right)=\frac{0+4}{0+\frac{3}{2}}=\frac{8}{3}\)
f/ \(lim\left(\frac{9^n-5.5^n+7.7^n}{9.3^n+5^n+2.8^n}\right)=lim\left(\frac{1-5.\left(\frac{5}{9}\right)^n+7\left(\frac{7}{9}\right)^n}{9.\left(\frac{1}{3}\right)^n+\left(\frac{5}{9}\right)^n+2.\left(\frac{8}{9}\right)^n}\right)=\frac{1}{0}=+\infty\)
g/ \(lim\left(\frac{6.6^n+3^5.9^n}{3^3.9^n-\frac{1}{2}.4^n}\right)=lim\left(\frac{6\left(\frac{2}{3}\right)^n+3^5}{3^3-\frac{1}{2}\left(\frac{4}{9}\right)^n}\right)=\frac{3^5}{3^3}=9\)
1.
\(\lim \frac{3n^2+5n+4}{2-n^2}=\lim \frac{\frac{3n^2+5n+4}{n^2}}{\frac{2-n^2}{n^2}}=\lim \frac{3+\frac{5}{n}+\frac{4}{n^2}}{\frac{2}{n^2}-1}=\frac{3}{-1}=-3\)
2.
\(\lim \frac{2n^3-4n^2+3n+7}{n^3-7n+5}=\lim \frac{\frac{2n^3-4n^2+3n+7}{n^3}}{\frac{n^3-7n+5}{n^3}}=\lim \frac{2-\frac{4}{n}+\frac{3}{n^2}+\frac{7}{n^3}}{1-\frac{7}{n^2}+\frac{5}{n^3}}=\frac{2}{1}=2\)
3.
\(\lim (\frac{2n^3}{2n^2+3}+\frac{1-5n^2}{5n+1})=\lim (n-\frac{3n}{2n^2+3}+\frac{1}{5}-n-\frac{1}{5n+1})\)
\(=\frac{1}{5}-\lim (\frac{3n}{2n^2+3}+\frac{1}{5n+1})=\frac{1}{5}-\lim (\frac{3}{2n+\frac{3}{n}}+\frac{1}{5n+1})=\frac{1}{5}-0=\frac{1}{5}\)
4.
\(\lim \frac{1+3^n}{4+3^n}=\lim (1-\frac{3}{4+3^n})=1-\lim \frac{3}{4+3^n}=1-0=1\)
5.
\(\lim \frac{4.3^n+7^{n+1}}{2.5^n+7^n}=\lim \frac{\frac{4.3^n+7^{n+1}}{7^n}}{\frac{2.5^n+7^n}{7^n}}\)
\(=\lim \frac{4.(\frac{3}{7})^n+7}{2.(\frac{5}{7})^n+1}=\frac{7}{1}=7\)
a/ \(=lim\frac{1}{\sqrt{n+1}+\sqrt{n}}=\frac{1}{\infty}=0\)
b/ \(=lim\frac{6n+1}{\sqrt{n^2+5n+1}+\sqrt{n^2-n}}=\frac{6+\frac{1}{n}}{\sqrt{1+\frac{5}{n}+\frac{1}{n^2}}+\sqrt{1-\frac{1}{n}}}=\frac{6}{1+1}=3\)
c/ \(=lim\frac{6n-9}{\sqrt{3n^2+2n-1}+\sqrt{3n^2-4n+8}}=lim\frac{6-\frac{9}{n}}{\sqrt{3+\frac{2}{n}-\frac{1}{n^2}}+\sqrt{3-\frac{4}{n}+\frac{8}{n^2}}}=\frac{6}{\sqrt{3}+\sqrt{3}}=\sqrt{3}\)
d/ \(=lim\frac{\left(\frac{2}{6}\right)^n+1-4\left(\frac{4}{6}\right)^n}{\left(\frac{3}{6}\right)^n+6}=\frac{1}{6}\)
e/ \(=lim\frac{\left(\frac{3}{5}\right)^n-\left(\frac{4}{5}\right)^n+1}{\left(\frac{3}{5}\right)^n+\left(\frac{4}{5}\right)^n-1}=\frac{1}{-1}=-1\)
f/ Ta có công thức:
\(1+3+...+\left(2n+1\right)^2=\left(n+1\right)^2\)
\(\Rightarrow lim\frac{1+3+...+2n+1}{3n^2+4}=lim\frac{\left(n+1\right)^2}{3n^2+4}=lim\frac{\left(1+\frac{1}{n}\right)^2}{3+\frac{4}{n^2}}=\frac{1}{3}\)
g/ \(=lim\left(\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{n}-\frac{1}{n+1}\right)=lim\left(1-\frac{1}{n+1}\right)=1-0=1\)
h/ Ta có: \(1^2+2^2+...+n^2=\frac{n\left(n+1\right)\left(2n+1\right)}{6}\)
\(\Rightarrow lim\frac{n\left(n+1\right)\left(2n+1\right)}{6n\left(n+1\right)\left(n+2\right)}=lim\frac{2n+1}{6n+12}=lim\frac{2+\frac{1}{n}}{6+\frac{12}{n}}=\frac{2}{6}=\frac{1}{3}\)
Đặt tổng là S n và tính 2 S n
ĐS: