1/65 < 1/5³ + 1/6³ + 1/7³ +...+ 1/n³ + 1/2004³ < 1/40
\(choA=\dfrac{1}{5^2}+\dfrac{1}{6^2}+\dfrac{1}{7^2}+...+\dfrac{1}{n^2}+...+\dfrac{1}{2004^2}\)
\(chứngminh\dfrac{1}{65}< A< \dfrac{1}{4}\)
\(A< \dfrac{1}{4.5}+\dfrac{1}{5.6}+\dfrac{1}{6.7}+...+\dfrac{1}{2003.2004}\)
\(\Rightarrow A< \dfrac{1}{4}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{6}+...+\dfrac{1}{2003}-\dfrac{1}{2004}\)
\(\Rightarrow A< \dfrac{1}{4}-\dfrac{1}{2004}< \dfrac{1}{4}\)
Đồng thời:
\(A>\dfrac{1}{5.6}+\dfrac{1}{6.7}+\dfrac{1}{7.8}+...+\dfrac{1}{2004.2005}\)
\(A>\dfrac{1}{5}-\dfrac{1}{6}+\dfrac{1}{6}-\dfrac{1}{7}+...+\dfrac{1}{2004}-\dfrac{1}{2005}\)
\(A>\dfrac{1}{5}-\dfrac{1}{2005}=\dfrac{80}{401}>\dfrac{50}{500}>\dfrac{1}{10}>\dfrac{1}{65}\)
Vậy \(\dfrac{1}{65}< A< \dfrac{1}{4}\)
Chứng minh rằng : 1/65 < 1/5^3 + 1/6^3 + 1/7^3 + ... + 1/2004^3 <1/40
cho \(A=\frac{1}{5^3}+\frac{1}{6^3}+\frac{1}{7^3}+....+\frac{1}{2004^3}\) và \(B=\frac{1}{65}\)
So sánh A và B
số 2 : tính nhanh
A = 99 - 97 + 95 -93 + 91 - 89 + .......... + 7 - 5 + 3 - 1
B = 50 - 49 + 48 - 47 + 46 - 45 + ........... + 4 - 3 + 2 - 1
C = 100 + 98 + 96 + ......... + 2 - 97 - 95 - ......... - 1
D = 1 + 3 + 5 + 7 + ......... + 999
E = 1 + 11 + 21 + 31 + ......... + 991
F = 3 + 7 + 11 + 15 + ......... + 99
H = 1 + 2 + 3 - 4 - 5 - 6 + 7 + 8 + 9 - 10 - 11 - 12 + .......... + 97 + 98 + 99 - 100 -101 - 102
I = 1 - 3 + 5 5 - 7 + 9 - 11 + ... + 2004 - 2007
K = -1 + 2 - 3 + 4 - 5 + 6 - 7 +.... + 2004 - 2005
G =1 - 2 - 3 + 4 + 5 - 6 - 7 + 8 +..... + 2004 + 2005
N = 1 - 4 + 7 - 10 + ..... + 2995 - 2998
Chứng minh rằng:
\(\frac{1}{65}\)<\(\frac{1}{5^3}\)+\(\frac{1}{6^3}\)+\(\frac{1}{7^3}\)+...+\(\frac{1}{2004^3}\)<\(\frac{1}{40}\)
tinh nhanh
A= 11+14+17+...+62+65
B=1/3 +1/15+1/35+1/63+1/99
C=3/10+3/40+3/88+3/154
D=3/5*7 + 3/7*9 +...+ 3/41*43 + 3/43*45
E=2003*2004/2003*2004+1 và 2004/2003
G=9*(151515/171717 + 131313/181818)
con nào làm đc thì làm trước nha. đang vội
\(A=11+14+17+...+62+65\)
Số số hạng của \(A\)là
\(\left(65-11\right)\div3+1=19\)(số hạng)
Tổng của \(A\)là:
\(\left(11+65\right)\times19\div2=722\)
Đáp số: 722
\(B=\frac{1}{3}+\frac{1}{15}+\frac{1}{35}+\frac{1}{63}+\frac{1}{99}\)
\(B=\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+\frac{1}{7.9}+\frac{1}{9.11}\)
\(B=\left(\frac{3-1}{1.3}+\frac{5-3}{3.5}+.....+\frac{9-7}{7.9}+\frac{11-9}{9.11}\right)\times\frac{1}{2}\)
\(B=\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+.....+\frac{1}{9}-\frac{1}{11}\right)\times\frac{1}{2}\)
\(B=\left(1-\frac{1}{11}\right)\times\frac{1}{2}\)
\(B=\frac{10}{11}\times\frac{1}{2}\)
\(B=\frac{5}{11}\)
\(C=\frac{3}{10}+\frac{3}{40}+\frac{3}{88}+\frac{3}{154}\)
\(C=\frac{3}{2.5}+\frac{3}{5.8}+\frac{3}{8.11}+\frac{3}{11.14}\)
\(C=\left(\frac{3}{2}-\frac{3}{5}+\frac{3}{5}-\frac{3}{8}+....+\frac{3}{11}-\frac{3}{14}\right)\div3\)
\(C=\left(\frac{3}{2}-\frac{3}{14}\right)\div3\)
\(C=\frac{9}{7}\div3\)
\(C=\frac{3}{7}\)
\(B=\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+\frac{1}{7.9}+\frac{1}{9.11}\)
\(B=1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+\frac{1}{9}-\frac{1}{11}\)
\(B=1-\frac{1}{11}\)
\(B=\frac{10}{11}\)
\(b=\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+\frac{1}{7.9}+\frac{1}{9.11}\)
\(b=1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+\frac{1}{9}-\frac{1}{11}\)
\(b=1-\frac{1}{11}\)
\(b=\frac{10}{11}\)
chứng minh rằng \(\frac{1}{65}\)<\(\frac{1}{5^3}+\frac{1}{6^3}+...+\frac{1}{2004^3}\)<\(\frac{1}{40}\)
a) 1 - 2 - 3 + 4 +5 - 6 - 7 + ..... + 2001 - 2002 -2003 + 2004
b) 1 + 2 - 3 - 4 + 5 + 6 - 7 - 8 + ..... + 2001 + 2002 - 2003 - 2004
a) \(1-2-3+4+5-6-7+...+2001-2002-2003+2004\)
\(=\left(1-2-3+4\right)+\left(5-6-7+8\right)+...+\left(2001-2002-2003+2004\right)\)
\(=0+0+...+0=0\)
b) \(1+2-3-4+5+6-7-8+...+2001+2002-2003-2004\)
\(=\left(1+2-3-4\right)+\left(5+6-7-8\right)+...+\left(2001+2002-2003-2004\right)\)
\(=\left(-4\right)+\left(-4\right)+...+\left(-4\right)\)
\(=\left(-4\right)\cdot501=\left(-2004\right)\)
Tính các tổng:
1/ S = 1 - 2 - 3 + 4 + 5 - 6 - 7 + 8 + ...+ 2001- 2002 - 2003 + 2004
2/ S = 1 + 2 - 3 - 4 + 5 + 6 - 7 - 8 + 9 + ...+ 2002 - 2003 - 2004 + 2005 + 2006
S=(1+2-3-4)+(5+6-7-8)+......+(2001+2002-2003-2004)+(2005+2006)
S=(-4)+(-4)+.......+(-4)+(2005+2006)
Dãy S có 2004-1:1+1=2004 số hạng
Dãy S có 2004:4=501 số -4
Do đó S=-4.501=-2004
S=-2004+(2005+2006)
S=-2004+4011
S=2007
1,S=(1-2-3+4)+(5-6-7+8)+.......+(2001-2002-2003+2004)
S=0+0+.........................+0
S=0
2,hình như pan gi sai đề