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 \(\frac{M}{N}\)

Tính :

1 + 3 + 5 + 7 + ... + (2n - 1) = 225 

Giải :

Theo công thức tính dãy số , ta có :

\(\frac{\left\{\left[\left(2n-1\right)-1\right]:2+1\right\}.\left[\left(2n-1\right)+1\right]}{2}=225\)

\(\frac{\left\{\left[2n-2\right]:2+1\right\}.2n}{2}=225\)

\(\left\{\left[2n-2\right]:2+1\right\}.n=450\)(Lượt giản thừa số 2)

\(\left\{\frac{2n-2}{2}+1\right\}.n=225\)

\(\left\{\frac{2n-2}{2}+\frac{2}{2}\right\}.n=225\)

\(\frac{2n-2+2}{2}.n=225\)

\(\frac{2n}{2}.n=225\)

\(n^2=225\)

\(\Rightarrow n=\sqrt{225}=15\)

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)\(\frac{1}{1.2.3}+\frac{1}{2.3.4}+\frac{1}{3.4.5}+...+\frac{1}{\left(n-1\right)n\left(n+1\right)}\)

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