Ta có :
\(\frac{1}{2!}+\frac{2}{3!}+\frac{3}{4!}+...+\frac{99}{100!}\)
\(=\frac{2-1}{2!}+\frac{3-1}{3!}+\frac{4-1}{4!}+...+\frac{100-1}{100!}\)
\(=\frac{1}{1!}-\frac{1}{2!}+\frac{1}{2!}-\frac{1}{3!}+\frac{1}{3!}-\frac{1}{4!}+...+\frac{1}{99!}-\frac{1}{100!}\)
\(=1-\frac{1}{100!}< 1\)
VẬY : \(\frac{1}{2!}+\frac{2}{3!}+\frac{3}{4!}+...+\frac{99}{100!}< 1\)
Tính Nhanh
\(\frac{\left(1+2+3+...+99+100\right).\left(\frac{1}{2}-\frac{1}{3}-\frac{1}{7}-\frac{1}{9}\right).\left(63.1,2-21.3,6\right)}{1-2+3-4+...+99-100}\)
CMR: \(\frac{1}{2!}+\frac{2}{3!}+\frac{3}{4!}+...+\frac{99}{100!}< 1\)
chứng minh rằng :
\(\frac{1}{3}+\frac{2}{3^2}+\frac{3}{3^3}+\frac{3}{3^4}+....+\frac{3}{3^{100}}< \frac{3}{21}\)
chứng minh rằng :
chứng minh rằng :
\(\frac{1}{3}+\frac{2}{3^2}+\frac{3}{3^3}+...+\frac{100}{3^{100}}< \frac{3}{4}\)
chứng minh rằng:
\(\frac{1.2-1}{2!}+\frac{2.3-1}{3!}+\frac{3.4-1}{4!}+...+\frac{99.100-1}{100!}< 2\)
mình ngu toán chúng minh (hép mi)
Chứng mình rằng:
\(\frac{1}{2^2}\)+\(\frac{1}{3^2}\)+\(\frac{1}{4^2}\)+...+\(\frac{1}{99^2}\) < 1
chứng minh rằng :
\(\frac{1}{3}+\frac{2}{3^2}+\frac{3}{3^3}+...+\frac{100}{3^{100}}< \frac{3}{4}\)
Cho M= 1-\(\frac{1}{2^2}\)-\(\frac{1}{3^2}\)-\(\frac{1}{4^2}\)-....-\(\frac{1}{100^2}\). CMR: M>\(\frac{1}{100}\)
Chứng minh rằng :
\(\frac{\frac{1}{2}+\frac{1}{4}+.....+\frac{1}{2n}}{\frac{1}{3}+\frac{1}{5}+.....+\frac{1}{2n-1}}< \frac{n}{n+1}\)
