1. Chứng minh : B = \(\left(1-\frac{2}{6}\right).\left(1-\frac{2}{12}\right).\left(1-\frac{2}{20}\right)...\left(1-\frac{2}{n\left(n+1\right)}\right)>\frac{1}{3}\)
2. cho M = \(\frac{1}{1.\left(2n-1\right)}+\frac{1}{3.\left(2n-3\right)}+\frac{1}{5.\left(2n-5\right)}+...+\frac{1}{\left(2n-3\right).3}+\frac{1}{\left(2n-1\right).1}\)
N = \(1+\frac{1}{3}+\frac{1}{5}+...+\frac{1}{2n-1}\)
Rút gọn \(\frac{M}{N}\)
Chứng minh
a) \(\frac{1}{2^2}+\frac{1}{4^2}+\frac{1}{6^2}+...+\frac{1}{\left(2n\right)^2}< \frac{1}{2}\)
b) \(\frac{1}{3^2}+\frac{1}{5^2}+\frac{1}{7^2}+...+\frac{1}{\left(2n+1\right)^2}< \frac{1}{4}\)
Chứng minh với \(n\in N\)\(n\ge1\)
Ta có a)\(\frac{1}{2^2}+\frac{1}{4^2}+\frac{1}{6^2}+...+\frac{1}{\left(2n\right)^2}< \frac{1}{2}\)
b)\(\frac{1}{3^2}+\frac{1}{5^2}+\frac{1}{7^2}+...+\frac{1}{\left(2n+1\right)^2}< \frac{1}{4}\)
Rút gọn:
B= \(\frac{1^2}{2^2-1}.\frac{3^2}{4^2-1}.\frac{5^2}{6^2-1}....\frac{\left(2n+1\right)^2}{\left(2n+2\right)^2-1}\)
help me! (ngu toàn tập)
a)\(\frac{3}{\left(1.2\right)^2}+\frac{5}{\left(2.3\right)^2}+...+\frac{2n+1}{\left[n\left(n+1\right)\right]^2}\)
b)\(\left(1-\frac{1}{2^2}\right)\left(1-\frac{1}{3^2}\right)\left(1-\frac{1}{4^2}\right)....\left(1-\frac{1}{n^2}\right)\)
c)\(\frac{150}{5.8}+\frac{150}{8.11}+\frac{150}{11.14}+...+\frac{150}{47.50}\)
d)\(\frac{1}{1.2.3}+\frac{1}{2.3.4}+\frac{1}{3.4.5}+...+\frac{1}{\left(n-1\right)n\left(n+1\right)}\)
Phân tích đa thức thành nhân tử :
\(\frac{\left(1^4+\frac{1}{4}\right)\left(3^4+\frac{1}{4}\right)\left(5^4+\frac{1}{4}\right)...\left(19^4+\frac{1}{4}\right)}{\left(2^4+\frac{1}{4}\right)\left(4^4+\frac{1}{4}\right)\left(6^4+\frac{1}{4}\right)...\left(20^4+\frac{1}{4}\right)}\)
Thầy tớ có cho gợi ý ạ :
Có dạng tổng quát của mỗi thừa số của tử số và mẫu số : \(n^4+\frac{1}{4}\)
Phân tích thành : \(n^4+\frac{1}{4}\)
\(=\left(n^2\right)^2+2n^2.\frac{1}{2}+\frac{1}{4}-n^2\)
\(=\left(n^2+\frac{1}{2}\right)^2-n^2\)
\(=\left(n^2-n+\frac{1}{2}\right)\left(n^2+n+\frac{1}{2}\right)\)
P/s : Giải giúp tớ nhé :33
CMR: \(\frac{1}{3^2}+\frac{1}{5^2}+\frac{1}{7^2}+...+\frac{1}{\left(2n+1\right)^2}< \frac{1}{4}\)
Chứng minh \(\frac{1}{4+1^4}+\frac{3}{4+3^4}+...+\frac{2n-1}{\left(4++\left(2n-1\right)\right)^4}=\frac{^{n^2}}{4n^2+1}\)
1/(4+1^4)+3/(4+3^4)+...+(2n-1)/(4+(2n-1)^4)=n^2/(4n^2+1)
tính :
B=\(\frac{1^2}{2^2-1}.\frac{3^2}{4^2-1}..........\frac{\left(2n+1\right)^2}{\left(2n+2\right)^2-1}\)