Violympic toán 7
Các câu hỏi tương tự
CMR : 1/3 - 2/3^2 + 3^3 - 4/3^4 + .... + 99/3^99 - 100/3^100 < 3/16
1 tính
a, S= \(3+3^2+3^3+....+3^{100}\)
b, M= \(\frac{1}{4}+\frac{1}{4^2}+.....+\frac{1}{4^{100}}\)
Cho biểu thức \(C=\dfrac{1}{3}-\dfrac{2}{3^2}+\dfrac{3}{3^3}-\dfrac{4}{3^4}+...+\dfrac{99}{3^{99}}-\dfrac{100}{3^{100}}\)
Chứng minh \(C< \dfrac{3}{16}\)
Tính :
1, A = \(\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+........+\dfrac{1}{100}\)
2, B = \(\dfrac{1}{2}+\dfrac{2}{3}+\dfrac{3}{4}+.........+\dfrac{99}{100}\)
tìm x biết (1/1*2 + 1/2*3 + 1/3*4 + ... + 1/99*100) - 2x =1/2
CMR:
a) \(\frac{1}{2!}+\frac{2}{3!}+\frac{3}{4!}+...+\frac{99}{100!}< 1\)
b) \(\frac{1.2-1}{2!}+\frac{2.3-1}{3!}+\frac{3.4-1}{4!}+...+\frac{99.100-1}{100!}< 2\)
TG
27 tháng 3 2020 lúc 8:36
1/ Chứng minh: \(C=\frac{1}{3}-\frac{2}{3^2}+\frac{3}{3^3}-\frac{4}{3^4}....+\frac{99}{3^{99}}-\frac{100}{3^{100}}\). Chứng minh: C < \(\frac{3}{16}\)
Tính
A = ( 1 - 1/1+2 ) ( 1 - 1/1+2+3 )( 1 - 1/1+2+3+4 ) ....... ( 1 - 1/1+2+3+....+99 )( 1 - 1/1+2+3+....+100 )
\(\dfrac{\left(13\dfrac{1}{4}-2\dfrac{5}{27}-10\dfrac{5}{6}\right).230\dfrac{1}{25}+46\dfrac{3}{4}}{\left(1\dfrac{3}{10}+\dfrac{10}{3}\right):\left(12\dfrac{1}{3}-14\dfrac{2}{7}\right)}\)
\(\dfrac{\left(1+2+3+...+99+100\right)\left(\dfrac{1}{2}-\dfrac{1}{3}-\dfrac{1}{7}-\dfrac{1}{9}\right)\left(63.1,2-21.3,6\right)}{1-2+3-4+.....+99-100}\)