Tính tổng A:
\(A=\frac{1}{1\times300}+\frac{1}{2\times301}+\frac{1}{3\times302}+...+\frac{1}{101\times400}\)
Những câu hỏi liên quan
\(A=\frac{1}{1\times300}+\frac{1}{2\times301}+...+\frac{1}{101\times400}\)
\(B=\frac{1}{1\times102}+\frac{1}{2\times103}+....+\frac{1}{298\times399}+\frac{1}{299\times400}\)
\(Tính:\frac{A}{B}\)
Giups mk voi mk can gap
A=\(\frac{1}{1.300}+\frac{1}{2.301}+\frac{1}{3.302}+...+\frac{1}{101.400}\)
299.A= 299.(\(\frac{1}{1.300}+\frac{1}{2.301}+\frac{1}{3.302}+...+\frac{1}{101.400}\))
299.A=\(\frac{299}{1.300}+\frac{299}{2.301}+\frac{299}{3.302}+...+\frac{299}{101.400}=\frac{1}{1}-\frac{1}{300}+\frac{1}{2}-\frac{1}{301}+...+\frac{1}{101}-\frac{1}{400}\)
A= \(=\frac{1}{299}\left(1+\frac{1}{2}+...+\frac{1}{101}-\frac{1}{300}-\frac{1}{301}-...-\frac{1}{400}\right)\)
Tương tự
B=\(\frac{1}{101}.\left(\frac{1}{1}-\frac{1}{102}+\frac{1}{2}-\frac{1}{103}+...+\frac{1}{299}-\frac{1}{400}\right)\)
B= \(\frac{1}{101}.\left(\frac{1}{1}+\frac{1}{2}...+\frac{1}{299}-\frac{1}{102}-\frac{1}{103}-...-\frac{1}{400}\right)\)
B= \(\frac{1}{101}.\left(\frac{1}{1}+\frac{1}{2}...+\frac{1}{102}+\frac{1}{103}+...+\frac{1}{299}-\frac{1}{102}-\frac{1}{103}-...-\frac{1}{400}\right)\)
B= \(\frac{1}{101}.\left(\frac{1}{1}+\frac{1}{2}...+\frac{1}{101}-\frac{1}{300}-\frac{1}{301}-...-\frac{1}{400}\right)\)
Hai dấu ngoặc ở biểu thức A và biểu thức B như nhau
Vậy \(A:B=\frac{1}{299}:\frac{1}{101}=\frac{101}{299}\)
Tính:
\(\frac{1}{2\times6}+\frac{1}{4\times9}+\frac{1}{6\times12}+...+\frac{1}{198\times300}\)
Đang cần gấp ạ.
Ta có :
1/2x6 + 1/4x9 + 1/6x12 +...+1/198 x 300
= 1/6x2 + 1/6x6 + 1/6x12 + ....+1/6x9900
= 1/6 x ( 1/2 + 1/6 + 1/ 12 +...+1/9900)
= 1/6 x (1/1x2 + 1/2x3 + 1/3x4+...+1/99x100)
=1/6x (1-1/2 + 1/2-1/3 + 1/3 - 1/4 + ....+1/99-1/100)
=1/6x(1-1/100)
=1/6 x 99/100
= 33/200
k cho mình nha , học tốt
Tính \(A=\frac{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{102}}{\frac{101}{1}+\frac{100}{2}+\frac{99}{3}+...+\frac{1}{101}}\)
Tính \(A=\frac{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{102}}{\frac{101}{1}+\frac{100}{2}+\frac{99}{3}+...+\frac{1}{101}}\)
\(A=\frac{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{102}}{\frac{101}{1}+\frac{100}{2}+\frac{99}{3}+...+\frac{1}{101}}\)
\(A=\frac{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{102}}{\left(\frac{100}{2}+1\right)+\left(\frac{99}{3}+1\right)+...+\left(\frac{1}{101}+1\right)+1}\)
\(A=\frac{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{102}}{\frac{102}{2}+\frac{102}{3}+...+\frac{102}{101}+\frac{102}{102}}\)
\(A=\frac{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{102}}{102.\left(\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{101}+\frac{1}{102}\right)}\)
\(A=\frac{1}{102}\)
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Tính các tổng sau:
a) \(\frac{1}{7}+\frac{1}{7^2}+\frac{1}{7^3}+...+\frac{1}{7^{100}}.\)
b) \(-\frac{4}{5}+\frac{4}{5^2}-\frac{4}{5^3}+...+\frac{4}{5^{200}}.\)
c)\(\frac{-1}{3}+\frac{1}{3^2}-\frac{1}{3^3}+...+\frac{1}{3^{100}}-\frac{1}{3^{101}}\)
Đăt A = \(\frac{1}{7}+\frac{1}{7^2}+\frac{1}{7^3}+......+\frac{1}{7^{100}}\)
\(\Rightarrow7A=1+\frac{1}{7}+\frac{1}{7^2}+.....+\frac{1}{7^{100}}\)
\(\Rightarrow7A-A=1-\frac{1}{7^{100}}\)
\(\Rightarrow6A=1-\frac{1}{7^{100}}\)
\(\Rightarrow A=\frac{1-\frac{1}{7^{100}}}{6}\)
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Tính Tổng
A=1+6+11+16+21+.....+101
B=1.2+2.3+3.4+....+98.99
C=\(1-\frac{1}{3}+\frac{1}{3^2}-\frac{1}{3^3}+.....+\frac{1}{3^{100}}\)
*) A=1+6+11+16+21+....+101
Dãy trên có: \(\left(101-1\right):5+1=21\)(số số hạng)
\(\Rightarrow A=\frac{\left(101+1\right)\cdot21}{2}=1071\)
*) Đặt C=\(1^2+2^2+3^2+....+98^2=1\cdot1+2\cdot2+3\cdot3+....+98\cdot98\)
\(\Rightarrow B-C=\left(1\cdot2+2\cdot3+3\cdot4+....+98\cdot99\right)-\left(1\cdot1+2\cdot2+3\cdot3+....+98\cdot98\right)\)
\(=\left(1\cdot2-1\cdot1\right)+\left(2\cdot3-2\cdot2\right)+\left(3\cdot4-3\cdot3\right)+.....+\left(98\cdot99-98\cdot98\right)\)
\(=1\left(2-1\right)+2\left(3-2\right)+3\left(4-3\right)+....+98\left(99-1\right)\)
\(=1\cdot1+2\cdot1+3\cdot1+....+98\cdot1\)
\(=1+2+3+....+98\)
\(=\frac{\left(98+1\right)\cdot98}{2}=4851\)
A = 1 + 6 + 11 + 16 +21 +... + 101
Số chữ số của tổng A là :
( 101 - 1 ) : 5 + 1 = 21 (số)
Tổng A = 1 + 6 + ... + 101 = (101 + 1) . 21 : 2 = 1071
B = 1.2 + 2.3 +3.4 + ... + 98.99
3B = 1.2.3 + 2.3.3 +... + 98.99.3
3B = 1.2.3 + 2.3.(4 - 1) + ... + 98.99.(100 - 97)
3B = 1.2.3 + 2.3.4 - 1.2.3 + ... + 98.99.100 - 97.98.99
3B = 98.99.100
B = 323400
\(C=1-\frac{1}{3}+\frac{1}{3^2}-\frac{1}{3^3}+...+\frac{1}{3^{100}}\)
\(3C=3-1+\frac{1}{3}-\frac{1}{3^2}+...+\frac{1}{3^{99}}\)
\(4C=3C+C=3+\frac{1}{3^{100}}\)
\(C=\frac{3^{101}+1}{4.3^{100}}\)
Xem thêm câu trả lời
A1+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{10...
Đọc tiếp
\(A=1+\frac{3}{2^3}+\frac{4}{2^4}+...+\frac{100}{2^{100}}\)
\(\frac{A}{2}=\frac{1}{2}+\frac{3}{2^4}+\frac{4}{2^5}+....+\frac{100}{2^{101}}\)\(A-\frac{A}{2}=\left(1+\frac{3}{2^3}+....+\frac{100}{2^{100}}\right)-\left(\frac{1}{2}+\frac{3}{2^4}+.....+\frac{100}{2^{101}}\right)\)
\(\frac{A}{2}=\frac{1}{2}+\frac{3}{2^3}+\frac{1}{2^4}+\frac{1}{2^5}+....+\frac{1}{2^{100}}-\frac{100}{2^{101}}\)
\(\frac{A}{2}=\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+\frac{1}{2^4}+....+\frac{1}{2^{100}}-\frac{1}{2^{101}}\)
\(\frac{A}{2}=\left(1-\left(\frac{1}{2}\right)^{101}\right).2-\frac{100}{2^{101}}\)
\(\frac{A}{2}=\frac{2^{101}-1}{2^{100}}-\frac{100}{2^{101}}\)
\(A=\frac{2^{101}-1}{2^{99}}-\frac{100}{2^{100}}\)
Tính A=\(\frac{\frac{1}{1\cdot300}+\frac{1}{2\cdot301}+\frac{1}{3\cdot302}+...+\frac{1}{101\cdot400}}{\frac{1}{1\cdot102}+\frac{1}{2\cdot103}+\frac{1}{3\cdot104}+...+\frac{1}{299\cdot400}}\)
Gọi tử số là \(C\)và mẫu số là \(D\)
Ta có:
\(A=\frac{C}{D}\)
\(C=\frac{1}{1.300}+\frac{1}{2.301}+\frac{1}{3.102}+...+\frac{1}{101.400}\)
\(C=\frac{1}{299}\left[\left(1-\frac{1}{300}\right)\right]+\left(\frac{1}{2}-\frac{1}{301}\right)+\left(\frac{1}{3}-\frac{1}{302}\right)+...+\left(\frac{1}{101}-\frac{1}{400}\right)\)
\(C=\frac{1}{299}\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{101}-\frac{1}{300}-\frac{1}{301}-\frac{1}{302}-...-\frac{1}{400}\right)\)
\(D=\frac{1}{1.102}+\frac{1}{2.103}+\frac{1}{3.104}+...+\frac{1}{299.400}\)
\(D=\frac{1}{101}\left[\left(1-\frac{1}{102}\right)+\left(\frac{1}{2}-\frac{1}{103}\right)+\left(\frac{1}{3}-\frac{1}{104}\right)+...+\left(\frac{1}{299}-\frac{1}{400}\right)\right]\)
\(D=\frac{1}{101}\left(1+\frac{1}{2}+\frac{1}{3}+..+\frac{1}{299}-\frac{1}{102}-\frac{1}{103}-\frac{1}{104}-...-\frac{1}{400}\right)\)
\(D=\frac{1}{101}\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{101}-\frac{1}{300}-\frac{1}{301}-\frac{1}{302}-...-\frac{1}{400}\right)\)
\(\Rightarrow A=\frac{C}{D}=\frac{\frac{1}{299}\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{101}-\frac{1}{300}-\frac{1}{301}-\frac{1}{302}-...-\frac{1}{400}\right)}{\frac{1}{101}\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{101}-\frac{1}{300}-\frac{1}{301}-\frac{1}{302}-...-\frac{1}{400}\right)}\)
\(=\frac{\frac{1}{299}}{\frac{1}{101}}=\frac{101}{299}.\)
Vậy \(A=\frac{101}{299}.\)
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1/ Cho Afrac{1}{1.102}+frac{1}{2.103}+...+frac{1}{299.400}Chứng minh rằng: Afrac{1}{101}left[left(1+frac{1}{2}+...+frac{1}{101}right)-left(frac{1}{300}+frac{1}{301}+...+frac{1}{400}frac{ }{ }right)right]2/ Tính Afrac{1}{2^2}+frac{1}{3^2}+...+frac{1}{2005^2}. Chứng minh A 13/ Cho Afrac{1}{101}+frac{1}{102}+...+frac{1}{200}Chứng minh: frac{1}{2} A 1GIÚP MÌNH NHA, MÌNH ĐANG CẦN GẤP.MÌNH SẼ TICK AI NHANH NHẤT!!
Đọc tiếp
1/ Cho \(A=\frac{1}{1.102}+\frac{1}{2.103}+...+\frac{1}{299.400}\)
Chứng minh rằng: \(A=\frac{1}{101}\left[\left(1+\frac{1}{2}+...+\frac{1}{101}\right)-\left(\frac{1}{300}+\frac{1}{301}+...+\frac{1}{400}\frac{ }{ }\right)\right]\)
2/ Tính \(A=\frac{1}{2^2}+\frac{1}{3^2}+...+\frac{1}{2005^2}\). Chứng minh \(A< 1\)
3/ Cho \(A=\frac{1}{101}+\frac{1}{102}+...+\frac{1}{200}\)
Chứng minh: \(\frac{1}{2}< A< 1\)
GIÚP MÌNH NHA, MÌNH ĐANG CẦN GẤP.MÌNH SẼ TICK AI NHANH NHẤT!!