(1- 1/2) . (1-1/3 ) ...........(1-1/2003) = ?
(1/2003+1/2004-1/2005)/(5/2003+5/2004-5/2005)-(2/2002+2/2003-2/2004)/(3/2002+3/2003-3/2004)
Chứng minh 1-1/2+1/3-1/4+...+1/2002-1/2003 = 1/1002+1/1003+...+1/2003
Đáp án của tớ là:
\(\frac{1}{1002}+\frac{1}{1003}+...+\frac{1}{2003}=\)\(\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2003}\right)-\)\(\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{1001}\right)\)
\(=\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2003}\right)-\)\(\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{6}+...+\frac{1}{2002}\right)-\)\(\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{6}+...+\frac{1}{2002}\right)=\)\(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2003}-\frac{1}{2}-\frac{1}{4}-\frac{1}{6}-...-\frac{1}{2002}\)\(-\frac{1}{2}-\frac{1}{4}-\frac{1}{6}-...-\frac{1}{2002}\)
Vậy:\(1+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{2003}=\frac{1}{1002}+\frac{1}{1003}+...+\frac{1}{2003}\)
xin chòa hôm nay mình sẽ giúp bạn lam bài toán này
ta có
1/1002+1/1003+....+1/2003=(1+1/2+1/3+.....+1/2003)-(1+1/2+1/3+....+1/1001)
1/1002+1/1003+....+1/2003=(1+1/2+1/3+.....+1/2003)-(1/2+1/4+1/6+....+1/2002)-(1/2+1/4+1/6+......+1/2002)
1/1002+1/1003+.....+1/2003=1+1/2+1/3+....+1/2003-1/2+1/4+1/6+....+1/2002-1/2-1/4-1/6-....-1/2002
Vậy1/1002+1/1002+.....+1/2003=1-1/2+1/3-1/4+....-2/2002-1/2003
Sửa: Vậy: \(1-\frac{1}{2}+\frac{1}{3}-...-\frac{1}{2003}=\frac{1}{1002}+\frac{1}{1003}+...+\frac{1}{2003}\)
Cho các số nguyên a^1;a^2;..;a^2003 thỏa mãn a^1+a^2+...+a^2003=0; a^1+a^2=a^3+a^4=...=a^2001+a^2002=a^2003+a^1=1.Tính a^1, a^2003
bài 1 : (4đ) 1) Tính : A = 1 phần 2003 + 1 phần 2004 - 1 phần 2005 : 5 phần 2003 + 5 phần 2004 - 5 phần 2005 - ( qua phân số khác rồi nhé ) 2/2002 + 2/2003 - 2/2004 : 3/2002 + 3/2003 - 3/2004 2) Cho B = 1/3+1/3 mũ 2 + 1/3 mũ 3 + 1/3 mũ 4 + ... +1/3 mũ 2015 + 1/3 mũ 2016 . Chứng minh ràng B<1/2
P=1/2003+1/2004-1/2004 - 2/2002+2/2003-2/2004
5/2003+5/2004-5/2005 3/2002+3/2003-3/2004
bài 1 : (4đ) 1) Tính : A = 1 phần 2003 + 1 phần 2004 - 1 phần 2005 : 5 phần 2003 + 5 phần 2004 - 5 phần 2005 - 2/2002 + 2/2003 - 2/2004 : 3/2002 + 3/2003 - 3/2004 2) Cho B = 1/3+1/3 mũ 2 + 1/3 mũ 3 + 1/3 mũ 4 + ... +1/3 mũ 2015 + 1/3 mũ 2016 . Chứng minh ràng B<1/2
nguyên một hàng mk đọc ko hỉu????????????
tinh A/B, biet
A=1/2*32+1/3*33+1/4*34+...+1/n*(n+30)+...+1/1973*2003
B=1/2*1974+1/3*1975+1/4*1976+...+1/n*(n+1972)+...+1/31*2003.
1/1*2+1/2*3+1/3*4....+1/2003*2004
\(\frac{1}{1\times2}+\frac{1}{2\times3}+\frac{1}{3\times4}+...+\frac{1}{2003\times2004}\)
=\(\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{2003}-\frac{1}{2004}\)
=\(\frac{1}{1}-\frac{1}{2004}=\frac{2004}{2004}-\frac{1}{2004}=\frac{2003}{2004}\)
A
a/ 1/1×2+1/2×3+1/3×4+...........+1/2003×200
Các bạn giúp mình nhé!! Đúng 22 giờ là mình phải nộp cho thầy rồi ạ!!!
\(\dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+....+\dfrac{1}{2003.200}\)
\(=1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+....+\dfrac{1}{2003}-\dfrac{1}{200}\)
\(=1-\dfrac{1}{200}\)
\(=\dfrac{199}{200}\)
tínhtổng 1\1*2+1\2*3+1\3*4+...+1\2003*2004
Đặt A=\(\frac{1}{1\cdot2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{2003.2004}\)
\(A=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{2003}-\frac{1}{2004}\)
\(A=1-\frac{1}{2004}\)
\(A=\frac{2003}{2004}\)
\(S=\frac{1}{1\cdot2}+\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+....+\frac{1}{2003\cdot2004}\)
\(S=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+....+\frac{1}{2003}-\frac{1}{2004}\)
\(S=1-\frac{1}{2004}\)
\(S=\frac{2003}{2004}\)