1) Cho a thỏa mãn: \(a^5-a^3+a=2\) Chứng minh rằng: \(a^6< 4\)
2) Chứng minh rằng: \(\frac{1^2}{1.3}+\frac{2^2}{3.5}+\frac{3^2}{5.7}+...+\frac{n^2}{\left(2n-1\right)\left(2n+1\right)}=\frac{n}{2}-\frac{n^2}{4n+2}\)
Những câu hỏi liên quan
17/lim\(\left(\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+...+\frac{1}{\left(2n-1\right)\left(2n+1\right)}\right)\)
18/lim\(\frac{1+a+a^2+...+a^n}{1+b+b^2+...+b^n}\left(\left|a\right|< 1;\left|b\right|< 1\right)\)
19/lim\(\frac{1-2+3-4+...+\left(2n-1\right)-2n}{2n+1}\)
cho \(I=\frac{1.3+2}{4}.\frac{3.5+2}{16}.....\frac{\left(2^{2n}-1\right)\left(2^{2n}+1\right)+2}{2^{2n}}\)với n thuộc N. chứng minh \(I< \frac{4}{3}\)
Cho M=\(\frac{1.3+2}{4}.\frac{3.5+2}{16}.\frac{15.17+2}{256}.\frac{255.257+2}{65536}.....\frac{\left(2^{2n}-1\right)\left(2^{2n}+1\right)+2}{2^{2n}}\)
(n thuộc N)
Chứng minh M<\(\frac{4}{3}\)
Cho \(A=\frac{1}{1.3}+\frac{1}{2.4}+\frac{1}{3.5}+\frac{1}{4.6}+...+\frac{1}{\left(2n-1\right)\left(2n+1\right)}+\frac{1}{2n\left(2n+2\right)}\)
a, Chứng minh \(A< \frac{3}{2}\) b, Tìm n để \(A=\frac{175}{132}\)
tự làm là hạnh phúc của mỗi công dân.
a)Tìm số nguyên dương n thỏa mãn:
\(\frac{1}{2}.\left(1+\frac{1}{1.3}\right).\left(1+\frac{1}{2.4}\right).\left(1+\frac{1}{3.5}\right)...\left(1+\frac{1}{n.\left(n+2\right)}\right)=\frac{2013}{2014}\)
b)tìm a sao cho
\(\left(a+\frac{1}{1.3}\right)+\left(a+\frac{1}{3.5}\right)+\left(a+\frac{1}{5.7}\right)+...+\left(a+\frac{1}{23.25}\right)=11.a+\left(\frac{1}{3}+\frac{1}{9}+\frac{1}{27}+\frac{1}{81}+\frac{1}{243}\right)\)
Chứng minh rằng:
1.2 + 2.3 + 3.4 +....+ n.(n+1) = \(\frac{n.\left(n+1\right).\left(n+2\right)}{3}\)
1.3 + 3.5 + 5.7 +.....+ n.(n+2)=\(\frac{3+n.\left(n+2\right).\left(n+4\right)}{6}\)
Giúp mk vs
Đặt \(A=1.2+2.3+3.4+...+n\left(n+1\right)\)
\(\Rightarrow3A=1.2.3+2.3.3+3.4.3+...+3n\left(n+1\right)\)
\(=1.2.3+2.3.\left(4-1\right)+3.4.\left(5-2\right)+...+n\left(n+1\right)\left(n+2-n+1\right)\)
\(=1.2.3+2.3.4-1.2.3+...+n\left(n+1\right)\left(n+2\right)-\left(n-1\right)n\left(n+1\right)\)
\(=n\left(n+1\right)\left(n+2\right)\)
\(\Rightarrow1.2+2.3+3.4+...+n\left(n+1\right)=\frac{n\left(n+1\right)\left(n+2\right)}{3}\)
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Bạn ơi tại sao 3n.(n+1) lại bằng với n.(n+1).(n+2-n+1)
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Chứng minh rằng:
a)\(\frac{1.3.5...39}{21.22.23...40}=\frac{1}{2^{20}}\)
b)\(\frac{1.3.5...\left(2n-1\right)}{\left(n+1\right)\left(n+2\right)\left(n+3\right)...2n}=\frac{1}{2^n}\)với n thuộc N*
a) Nhân cả tử và mẫu với 2 . 4 . 6 ... 40 ta được :
\(\frac{1.3.5...39}{21.22.23...40}=\frac{\left(1.3.5...39\right).\left(2.4.6...40\right)}{\left(21.22.23...40\right).\left(2.4.6...40\right)}\)
\(=\frac{1.2.3...39.40}{1.2.3...40.2^{20}}=\frac{1}{2^{20}}\)
b) Nhân cả tử và mẫu với 2 . 4 . 6 ... 2n ta được :
\(\frac{1.3.5...\left(2n-1\right)}{\left(n+1\right)\left(n+2\right)\left(n+3....2n\right)}=\frac{1.3.5...\left(2n-1\right).\left(2.4.6...2n\right)}{\left(n+1\right)\left(n+2\right)...\left(2n\right).\left(2.4.6...2n\right)}\)
\(=\frac{1.2.3...\left(2n-1\right).2n}{1.2.3...2n.2^n}=\frac{1}{2^n}\)
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a/Chứng minh rằng \(\frac{2}{\left(2n+1\right)\sqrt{n}+\sqrt{n+1}}=\frac{1}{\sqrt{n}}-\frac{1}{\sqrt{n+1}}\)
b/Áp dụng chứng minh
\(\frac{1}{3\left(1+\sqrt{2}\right)}+\frac{1}{5\left(\sqrt{2}+\sqrt{3}\right)}+\frac{1}{7\left(\sqrt{3}+\sqrt{4}\right)}+...+\frac{1}{4003\left(\sqrt{2001}+\sqrt{2002}\right)}<\frac{2001}{2003}\)
Cho \(A=\frac{1}{4^2}+\frac{1}{6^2}+\frac{1}{8^2}+...+\frac{1}{\left(2n\right)^2}\left(n\in N,n.2\right)\)
Chứng minh A<1/4
Ta có :
\(A=\frac{1}{4^2}+\frac{1}{6^2}+\frac{1}{8^2}+...+\frac{1}{\left(2n\right)^2}\)
\(A=\frac{1}{2^2}\left(\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{n^2}\right)< \frac{1}{2^2}\left(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{\left(n-1\right)n}\right)\)
\(A< \frac{1}{4}\left(\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{n-1}-\frac{1}{n}\right)=\frac{1}{4}\left(1-\frac{1}{n}\right)\)
\(A< \frac{1}{4}-\frac{1}{4n}\)
Lại có \(n>0\) nên \(\frac{1}{4n}>0\)
\(\Rightarrow\)\(\frac{1}{4}-\frac{1}{4n}< \frac{1}{4}\)
Vậy \(A< \frac{1}{4}\)
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