1. Chứng minh răng với mọi số nguyên nguyên dương n ta luôn có
1/6 + 1/12 + 1/20 + …+ 1/n^2+3n+2 < 1/2
2. tính tổng S=1/1.2 + 1/2.3 + 1/3.4 + ... + 1/99.100
giúp với ạ
1. Chứng minh răng với mọi số nguyên nguyên dương n ta luôn có 1/6 + 1/12 + 1/20 + …+ 1/n^2+3n+2 < 1/2 2. tính tổng S=1/1.2 + 1/2.3 + 1/3.4 + ... +
By Adalynn
Bài 1:
Ta có:
$\frac{1}{n^2+3n+2}=$ $\frac{1}{(n+1)(n+2)}=$ $\frac{1}{n+1}-$ $\frac{1}{n+2}$
$\frac{1}{6}=$ $\frac{1}{2.3}=$ $\frac{1}{2}-$ $\frac{1}{3}$
$\frac{1}{12}=$ $\frac{1}{3.4}=$ $\frac{1}{3}-$ $\frac{1}{4}$
$\frac{1}{20}=$ $\frac{1}{4.5}=$ $\frac{1}{4}-$ $\frac{1}{5}$
$\frac{1}{n^2+3n+2}=$ $\frac{1}{(n+1)(n+2)}=$ $\frac{1}{n+1}+$ $\frac{1}{n+2}$
$\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+…+\frac{1}{n^2+3n+2}=\frac{1}{2}-\frac{1}{n+2}<\frac{1}{2}$
Bài 2:
Ta có: $\frac{1}{a(a+1)}=\frac{1}{a}-\frac{1}{a+1}$
⇒ $S=(\frac{1}{1}-\frac{1}{2})+(\frac{1}{2}-\frac{1}{3})+(\frac{1}{3}-\frac{1}{4})+…+(\frac{1}{99}-$$\frac{1}{100})=1-\frac{1}{100}=$ $\frac{99}{100}$
1, $\frac{1}{6}+$ $\frac{1}{12}+…+$ $\frac{1}{n^2+3n+2}$
$=\frac{1}{2.3}+$ $\frac{1}{3.4}+…+$ $\frac{1}{(n+1)(n+2)}$
$=\frac{1}{2}-$ $\frac{1}{3}+$ $\frac{1}{3}-$ $\frac{1}{4}+…+$ $\frac{1}{n+1}-$ $\frac{1}{n+2}$
$=\frac{1}{2}-$ $\frac{1}{n+2}<$ $\frac{1}{2}$
2, $\frac{1}{1.2}+$ $\frac{1}{2.3}+…+$ $\frac{1}{99.100}$
$=1-\frac{1}{2}+$ $\frac{1}{2}-$ $\frac{1}{3}+…+$ $\frac{1}{99}-$ $\frac{1}{100}$ $=1-\frac{1}{100}$
$=\frac{99}{100}$