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)

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

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

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

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

\(n^2=225\)

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

Câu 1:\(\frac{1}{3}+\frac{1}{15}+\frac{1}{35}+...+\frac{1}{^{n^2+2n}}\)(n\(\varepsilon\)N*)

Câu 2: \(\frac{1}{2}+\frac{1}{14}+\frac{1}{35}+...+\frac{1}{\left(n-3\right)\cdot n}\)

Câu 3: \(\frac{1}{90}-\frac{1}{72}-\frac{1}{42}-\frac{1}{30}-\frac{1}{20}-\frac{1}{6}-\frac{1}{2}\)

Câu 4: \(\left(1-\frac{1}{12}\right)+\left(1-\frac{1}{2\cdot3}\right)+\left(1-\frac{1}{3\cdot4}\right)+...+\left(1-\frac{1}{1995\cdot1996}\right)\)

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