Rút gọn:
\(p=\frac{\frac{1}{1\left(2n-1\right)}+\frac{1}{3\left(2n-3\right)}+\frac{1}{5\left(2n-5\right)}+...+\frac{1}{\left(2n-3\right)3}+\frac{1}{\left(2n-1\right)1}}{1+\frac{1}{3}+\frac{1}{5}+\frac{1}{2n-1}}\)
~Help mik vs~
\(\sin^3\frac{x}{3}+3\sin^3\frac{x}{3^2}+...+3^{n-1}\sin^3\frac{x}{3}=\frac{1}{4}\left(3^n\sin^3\frac{x}{3^n}-\sin x\right)\)\(\frac{1}{2}.\frac{3}{4}.\frac{5}{6}...\frac{2n+1}{2n+2}<\frac{1}{\sqrt{3n+4}}\left(n\ge1\right)\)\(\left(n!\right)^2\ge n^2\ge\left(n+1\right)^{n-1}cho\left(n\ge1\right)\)Chứng minh: \(\frac{1}{3\left(1+\sqrt{2}\right)}+\frac{1}{5\left(\sqrt{2}+\sqrt{3}\right)}+\frac{1}{7\left(\sqrt{3}+\sqrt{4}\right)}+...+\frac{1}{\left(2n+1\right)\left(\sqrt{n}+\sqrt{n+1}\right)}\)\(< \frac{1}{2}\)
Tính tổng:
\(S=\frac{3}{1^2.3}+\frac{5}{\left(1^2+2^2\right).4}+\frac{7}{\left(1^2+2^2+3^2\right).5}+...+\)\(\frac{2n+1}{\left(1^2+2^2+3^2+...+n^2\right).\left(n+2\right)}\)
mấy Box Toán giúp em với
cho f(n)=(n2 + n +1 )2 +1 với n thuộc N* . Đặt \(p_n=\frac{f_{\left(1\right)}\cdot f_{\left(3\right)}\cdot f_{\left(5\right)}\cdot\cdot\cdot\cdot\cdot\cdot\cdot\cdot f_{\left(2n-1\right)}}{f_{\left(2\right)}\cdot f_{\left(4\right)}\cdot f_{\left(6\right)}\cdot\cdot\cdot\cdot\cdot\cdot\cdot\cdot f_{\left(2n\right)}}\)
chứng minh rằng : P1 + P2 +P3 +................+ Pn <1/2
cho: \(n\in N\)
Cmr: \(\left(n^2+2n+5\right)^3-\left(n+1\right)^2+2012⋮6\)
CMR:
M=\(\frac{1}{3.\left(\sqrt{1}+\sqrt{2}\right)}\)+\(\frac{1}{5.\left(\sqrt{2}+\sqrt{3}\right)}\) +...+\(\frac{1}{\left(2n+1\right).\left(\sqrt{n}+\sqrt{n+1}\right)}< \frac{1}{2}\)
y=\(2x-1\left(d_1\right)\)
y=\(\left(2n-1\right)x+\dfrac{3}{2}\left(d_2\right)\)
y=\(-x+3\left(d_3\right)\)
Tìm n đồng quy
chứng minh
\(1^2+2^2+3^2+...+n^2=\frac{n\left(n+1\right)\left(2n+1\right)}{6}\)