Đặt \(A=\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+...+\frac{1}{3^{99}}\)
\(\Rightarrow3A=1+\frac{1}{3}+\frac{1}{3^2}+...+\frac{1}{3^{98}}\)
\(3A-A=\left(1+\frac{1}{3}+\frac{1}{3^2}+...+\frac{1}{3^{98}}\right)-\left(\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+...+\frac{1}{3^{99}}\right)\)
\(2A=1-\frac{1}{3^{99}}\)
\(\Rightarrow A=\frac{1-\frac{1}{3^{99}}}{2}< \frac{1}{2}\left(đpcm\right)\)
CMR : 1/3 - 2/3^2 + 3^3 - 4/3^4 + .... + 99/3^99 - 100/3^100 < 3/16
CMR:
a) \(\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+...+\frac{1}{3^{99}}< \frac{1}{2}\)
b) \(\frac{3}{1^2.2^2}+\frac{5}{2^2.3^2}+\frac{7}{3^2.4^2}+...+\frac{19}{9^2.10^2}< 1\)
CMR: \(C=\dfrac{1}{3}+\dfrac{2}{3^2}+\dfrac{3}{3^3}+...+\dfrac{99}{3^{99}}+\dfrac{100}{3^{100}}< \dfrac{3}{4}\)
CMR:
a) \(\frac{1}{2!}+\frac{2}{3!}+\frac{3}{4!}+...+\frac{99}{100!}< 1\)
b) \(\frac{1.2-1}{2!}+\frac{2.3-1}{3!}+\frac{3.4-1}{4!}+...+\frac{99.100-1}{100!}< 2\)
Tính tổng
1,A=a1+a2+...+an với ai-ai-1=k
2,B=1/1*2*3+1/2*3*4+...+1/48*49*50
3,C=1+1/3+1/5+...+1/97+1/99
/
1/1*99+1/3*97+...+1/97*3+1/99*1
chứng minh rằng 1/3+1/3^2+1/3^3+...+1/3^99<1/2
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 )
tính :1*2-1/2!-3*2-1/3!+3*4-1/4!+...+99*100-1/100!
Bài 1: Cho A=\(2+2^2+2^3+...+2^{60}\) CMR: \(A⋮3\) \(A⋮7\) \(A⋮15\)
Bài 2: Cho B=\(\dfrac{1}{2}+\left(\dfrac{1}{2}\right)^2+\left(\dfrac{1}{2}\right)^3+...+\left(\dfrac{1}{2}\right)^{98}+\left(\dfrac{1}{2}\right)^{99}\)
CMR: B<1
Bài 3: Tìm a,b,c\(\in Q\) biết: \(a.b=2\) ; \(b.c=3\) ; \(c.a=54\)
\(\dfrac{\left(13\dfrac{1}{4}-2\dfrac{5}{27}-10\dfrac{5}{6}\right).230\dfrac{1}{25}+46\dfrac{3}{4}}{\left(1\dfrac{3}{10}+\dfrac{10}{3}\right):\left(12\dfrac{1}{3}-14\dfrac{2}{7}\right)}\)
\(\dfrac{\left(1+2+3+...+99+100\right)\left(\dfrac{1}{2}-\dfrac{1}{3}-\dfrac{1}{7}-\dfrac{1}{9}\right)\left(63.1,2-21.3,6\right)}{1-2+3-4+.....+99-100}\)
