A = 1 + 2 + 2^2 + 2^3 + ... + 2^100
Rút gọn
A= 2^100+2^99+2^98.....+2+1
B=3^100+3^99+3^98....+3+1
C=4^100+4^99+....+4+1
D=2^100- 2^99+....+2^2 - 2 + 1
E=3^100 - 3^99 + 3^98....- 3 +1
Thu gọn
M= 2 + 2^2 + 2^3 ....+ 2^100
Cho A =2+2^2+2^3+....2^100. Tìm số tự nhiên x sao cho A + 1 = 2x
Bài 1:
a: \(2A=2^{101}+2^{100}+...+2^2+2\)
\(\Leftrightarrow A=2^{100}-1\)
b: \(3B=3^{101}+3^{100}+...+3^2+3\)
\(\Leftrightarrow2B=3^{100}-1\)
hay \(B=\dfrac{3^{100}-1}{2}\)
c: \(4C=4^{101}+4^{100}+...+4^2+4\)
\(\Leftrightarrow3C=4^{101}-1\)
hay \(C=\dfrac{4^{101}-1}{3}\)
A=1/2+2/2^2+3/2^3+...+100/2^100
B=1/3+2/3^2+3/3^2+...+100/3^100
\(A=1+\frac{3}{2^3}+\frac{4}{2^4}+...+\frac{100}{2^{100}}\)
\(\frac{A}{2}=\frac{1}{2}+\frac{3}{2^4}+\frac{4}{2^5}+....+\frac{100}{2^{101}}\)\(A-\frac{A}{2}=\left(1+\frac{3}{2^3}+....+\frac{100}{2^{100}}\right)-\left(\frac{1}{2}+\frac{3}{2^4}+.....+\frac{100}{2^{101}}\right)\)
\(\frac{A}{2}=\frac{1}{2}+\frac{3}{2^3}+\frac{1}{2^4}+\frac{1}{2^5}+....+\frac{1}{2^{100}}-\frac{100}{2^{101}}\)
\(\frac{A}{2}=\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+\frac{1}{2^4}+....+\frac{1}{2^{100}}-\frac{1}{2^{101}}\)
\(\frac{A}{2}=\left(1-\left(\frac{1}{2}\right)^{101}\right).2-\frac{100}{2^{101}}\)
\(\frac{A}{2}=\frac{2^{101}-1}{2^{100}}-\frac{100}{2^{101}}\)
\(A=\frac{2^{101}-1}{2^{99}}-\frac{100}{2^{100}}\)
Tính tổng: A=2^2 + 3^2 + 4^2 + ... + 100^2
A=2^2 + 3^2 + 4^2 + ... + 100^2
A=(2.3.4...100)+(2.3.4...100)+1
A=2(2.3.4....100)+1
A= ...
Tính tiếp, ngang đó mik chịu
Tính tổng: A=2^2 + 3^2 + 4^2 + ... + 100^2
A=2^2 + 3^2 + 4^2 + ... + 100^2
A=(2.3.4...100)+(2.3.4...100)+1
A=2(2.3.4....100)+1
A= ...
Tính tiếp, ngang đó mik chịu
1.A=2^2+4^2+6^2+...+100^2
2.A=1^2+2^2+3^2+...+99^2
3.A=1^2+2^2+3^2+...+100^2
CMR:
a)1/10^2 +1/11^2+1/12^2+...+1/100^2 >3/4
b)1/2^2+1/3^2+1/4^2+...+1/100^2<99/100
c)1/2^2+1/3^2+1/4^2+...+1/100^2<3/4
1/ Cho A= \(\dfrac{1}{3}\)-\(\dfrac{2}{3^2}\)+\(\dfrac{3}{3^3}\)-\(\dfrac{4}{3^4}\)+.....+\(\dfrac{99}{3^{99}}\)-\(\dfrac{100}{3^{100}}\) Chứng minh A < \(\dfrac{3}{16}\)
2/ Cho B=(\(\dfrac{1}{2^2}\)-1)(\(\dfrac{1}{3^2}\)-1)....(\(\dfrac{1}{100^2}\)-1) So sánh B và \(\dfrac{-1}{2}\)
2:
\(B=\left(\dfrac{1}{2^2}-1\right)\left(\dfrac{1}{3^2}-1\right)\cdot...\cdot\left(\dfrac{1}{100^2}-1\right)\)
\(=\left(\dfrac{1}{2}-1\right)\left(\dfrac{1}{2}+1\right)\left(\dfrac{1}{3}-1\right)\left(\dfrac{1}{3}+1\right)\cdot...\cdot\left(\dfrac{1}{100}-1\right)\left(\dfrac{1}{100}+1\right)\)
\(=\left(\dfrac{1}{2}-1\right)\left(\dfrac{1}{3}-1\right)\cdot...\cdot\left(\dfrac{1}{100}-1\right)\left(\dfrac{1}{2}+1\right)\left(\dfrac{1}{3}+1\right)\cdot...\cdot\left(\dfrac{1}{100}+1\right)\)
\(=\dfrac{-1}{2}\cdot\dfrac{-2}{3}\cdot...\cdot\dfrac{-99}{100}\cdot\dfrac{3}{2}\cdot\dfrac{4}{3}\cdot...\cdot\dfrac{101}{100}\)
\(=-\dfrac{1}{100}\cdot\dfrac{101}{2}=\dfrac{-101}{200}< -\dfrac{100}{200}=-\dfrac{1}{2}\)
Tính
A=1.(100-1)+2.(100-2)+3.(100-3)+..............+99.(100-99)
B=1.(100+1)+2.(100+2)+3.(100+3)+...........+99.(100+99)
Thu gọn tổng sau:
a) A=1+3+3^2+...+3^100
b) B=2^100-2^99+2^98-2^97+...+2^2-2
c) C=3^100-3^99+3^98-3^97+...+3^2-3+1
a) A =1+3+32+33+...+3100
3A = 3 + 32+33+...+3101
3A-A=( 3 + 32+33+...+3101)-(1+3+32+33+...+3100)
2A = 3101-1
A = \(\frac{3^{101}-1}{2}\)
Thùy An làm sai rùi
a) A=1+3+3^2+...+3^100
3A=3+3^2+....+3^101
3A-A=1+3^101
A=(1+3^101)/2
a) A=1+3+32+...+3100
3A= 3+32+...+3100+3101
3A-A=3101-1
2A=3101-1
A=(3101-1):2