3/Tìm số tự nhiên n sao cho
A= \(\frac{1}{1.3}\)+ \(\frac{1}{2.4}\)+ \(\frac{1}{3.5}\)+....+ \(\frac{1}{\left(2n-1\right).\left(2n+1\right)}\)+ \(\frac{1}{2n.\left(2n+2\right)}\)= \(\frac{14651}{19800}\)
4/ CMR: Nếu (n,6)=1 thì (n2-1) chia hết cho 24
Những câu hỏi liên quan
Rút gọn :\(A=\frac{1}{1.3}+\frac{1}{2.4}+\frac{1}{3.5}+\frac{1}{4.6}+...+\frac{1}{\left(2n-1\right)\left(2n+1\right)}+\frac{1}{2n\left(2n+2\right)}\)
tự làm là hạnh phúc của mỗi công dân.
Tính Q=\(\frac{1.3}{3.5}+\frac{2.4}{5.7}+\frac{3.5}{7.9}+.....+\frac{\left(n-1\right)\left(n+1\right)}{\left(2n-1\right)\left(2n+1\right)}+......+\frac{1002.1004}{2005.2007}\)
Tính: \(Q=\frac{1.3}{3.5}+\frac{2.4}{5.7}+\frac{3.5}{7.9}+...+\frac{\left(n-1\right)\left(n+1\right)}{\left(2n-1\right)\left(2n+1\right)}+...+\frac{1002.1004}{2005.2007}\)
Tính: \(Q=\frac{1.3}{3.5}+\frac{2.4}{5.7}+\frac{3.5}{7.9}+...+\frac{\left(n-1\right)\left(n+1\right)}{\left(2n-1\right)\left(2n+1\right)}+...+\frac{1002.1004}{2005.2007}\)
hơi khó đó tick mình nha Hoàng Thu Hà
Đúng 0
Bình luận (0)
CMR với mọi số tự nhiên n2 thì :a)frac{1}{2^2}+frac{1}{4^2}+frac{1}{6^2}+...+frac{1}{left(2nright)^2}frac{1}{2}b)frac{1}{3^2}+frac{1}{5^2}+frac{1}{7^2}+...+frac{1}{left(2n+1right)^2}frac{1}{4}c)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}{left(2n+1right)^2}right)2
Đọc tiếp
CMR với mọi số tự nhiên n>2 thì :
a)\(\frac{1}{2^2}+\frac{1}{4^2}+\frac{1}{6^2}+...+\frac{1}{\left(2n\right)^2}\)<\(\frac{1}{2}\)
b)\(\frac{1}{3^2}+\frac{1}{5^2}+\frac{1}{7^2}+...+\frac{1}{\left(2n+1\right)^2}\)<\(\frac{1}{4}\)
c)\(\left(1+\frac{1}{1.3}\right)\left(1+\frac{1}{2.4}\right)\left(1+\frac{1}{3.5}\right)...\left(1+\frac{1}{\left(2n+1\right)^2}\right)\)<2
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.
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\(\frac{1-2+3-4+...+\left(2n-1\right)-2n}{2n+1}\)
tính tổng:
I=\(\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+...+\frac{1}{\left(2n+1\right).\left(2n+3\right)}\)
\(I=\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+...+\frac{1}{\left(2n+1\right).\left(2n+3\right)}\)
\(\Rightarrow I=\frac{1}{2}\left(\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+...+\frac{2}{\left(2n+1\right).\left(2n+3\right)}\right)\)
\(\Rightarrow I=\frac{1}{2}\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{2n+1}-\frac{1}{2n+3}\right)\)
\(\Rightarrow I=\frac{1}{2}\left(1-\frac{1}{2n+3}\right)\)
\(\Rightarrow I=\frac{1}{2}.\frac{2n+2}{2n+3}\)
\(\Rightarrow I=\frac{n+1}{2n+3}\)
Đúng 0
Bình luận (0)
\(I=\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+...+\frac{1}{\left(2n+1\right)\left(2n+3\right)}=\frac{1}{2}.\left(\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+...+\frac{2}{\left(2n+1\right)\left(2n+3\right)}\right)\)
\(=\frac{1}{1}-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{2n+1}-\frac{1}{2n+3}=\frac{1}{1}-\frac{1}{2n+3}\)
\(=\frac{2n+3}{2n+3}-\frac{1}{2n+3}=\frac{2n+2}{2n+3}\)
Đúng 0
Bình luận (0)
Tính \(\lim\limits\left[\frac{1}{1.3}+\frac{1}{3.5}+...+\frac{1}{\left(2n-1\right)\left(2n+1\right)}\right]\)
\(=lim\frac{1}{2}\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+...+\frac{1}{2n-1}-\frac{1}{2n+1}\right)\)
\(=lim\frac{1}{2}\left(1-\frac{1}{2n+1}\right)=\frac{1}{2}.\left(1-0\right)=\frac{1}{2}\)
Đúng 0
Bình luận (0)