Tính tổng:
A= 5/1.4+5/4.7+...+5/100.103
B= 1/15+1/35+...+1/2499
Những câu hỏi liên quan
tính giá trị biểu thức
C=1/15+1/35+.................+1/2499
B=5/1.4+5/4.7+...................5/100.103
chứng minh rằng: a) a/n.n(n+a)=1/n-1/n+a ; b) áp dụng câu a tính: A=1/2.3+1/3.4+...+1/99.101 ; B=5/1.4+5/4.7+...+5/100.103 ; C=1/15+1/35+...+1/2499
tính tổng :
A = 5/1.4 + 1/4.7 + ...+5/100.103
B= 1/15 + 1/35 + ... +1/2499
C= 32/8.11 + 32/11.14 32/14.17 + ... +32197.20
Tính:
a) \(\frac{5}{1.4}+\frac{5}{4.7}+...+\frac{5}{100.103}\)
b) \(\frac{1}{15}+\frac{1}{35}+...+\frac{1}{2499}\)
a)\(\frac{5}{1.4}+\frac{5}{4.7}+...+\frac{5}{100.103}=\frac{5}{3}\cdot\left(\frac{3}{1.4}+\frac{4}{4.7}+...+\frac{3}{100.103}\right)\)
\(=\frac{5}{3}\cdot\left(1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+...+\frac{1}{100}-\frac{1}{103}\right)=\frac{5}{3}\cdot\left(1-\frac{1}{103}\right)=\frac{5}{3}\cdot\frac{102}{103}=\frac{170}{103}\)b)\(\frac{1}{15}+\frac{1}{35}+...+\frac{1}{2499}=\frac{1}{3.5}+\frac{1}{5.7}+...+\frac{1}{49.51}\)
\(\frac{1}{2}\cdot\left(\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{49}-\frac{1}{51}\right)=\frac{1}{2}\cdot\left(\frac{1}{3}-\frac{1}{51}\right)=\frac{1}{2}\cdot\frac{16}{51}=\frac{8}{51}\)
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Câu a) bạn Ác Mộng làm rồi nên mình làm b) nha
b)Gọi A = \(\frac{1}{15}+\frac{1}{35}+...+\frac{1}{2499}\)
\(A=\frac{1}{3.5}+\frac{1}{5.7}+...+\frac{1}{49.51}\)
\(2A=2.\left(\frac{1}{3.5}+\frac{1}{5.7}+...+\frac{1}{49.51}\right)\)
\(2A=\frac{2}{3.5}+\frac{2}{5.7}+...+\frac{2}{49.51}\)
\(2A=\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{49}-\frac{1}{51}\)
\(2A=\frac{1}{3}-\frac{1}{51}\)
\(2A=\frac{16}{51}\)
\(A=\frac{16}{51}:2\)
\(A=\frac{8}{51}\)
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5*(5-5/4+5/4-5/7+.......+5/100-5/103)
5*(5-5/103)
5*......... bạn tự tính nhé
câu b 1/3*5+1/5*7+............+1/49*51
1*(1/1-1/3+1/3-1/5+............+1/49-1/51)
1/1-1/51 tính ra rồi ra kết quả
tk nha
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Xem thêm câu trả lời
tính tổng
A= 5/1.4 + 5/4.7 +...+ 5/100.103
B= 1/15 + 1/35 +...+ 1/2499
C= 32 /8.11 + 32/ 11.14+ 32/14.17 + ...+ 32/197.20
GIÚP MIK VỚI. CẢM ƠN CÁC BẠN NHÌU NHA
_ Chứng minh rằng
a .
a / n ( n+ a ) = 1/ n - 1 / n + a ( n , a thuộc N * )
_ dấu / bằng chữ phần
b. Áp dụng câu a tính
A = 1/23 + 1/34 + ... + 1/99.100
B = 5/1.4 + 5 / 4.7 +.... + 5/100.103
C = 1/15 + 1/35 + ... + 1/2499
tính tổng: A dfrac{2}{1.3}+dfrac{2}{3.5}+dfrac{2}{5.7}+...+dfrac{2}{99.101} B dfrac{5}{1.3}+dfrac{5}{3.5}+dfrac{5}{3.7}+...+dfrac{5}{99.101}
C dfrac{1}{2.3}+dfrac{1}{3.4}+...+dfrac{1}{99.100} D dfrac{5}{1.4}+dfrac{5}{4.7}+...+dfrac{5}{100.103} E dfrac{1}{15}+dfrac{1}{35}+...+dfrac{1}{2499}
Đọc tiếp
tính tổng: A= \(\dfrac{2}{1.3}+\dfrac{2}{3.5}+\dfrac{2}{5.7}+...+\dfrac{2}{99.101}\) B= \(\dfrac{5}{1.3}+\dfrac{5}{3.5}+\dfrac{5}{3.7}+...+\dfrac{5}{99.101}\)
C= \(\dfrac{1}{2.3}+\dfrac{1}{3.4}+...+\dfrac{1}{99.100}\) D= \(\dfrac{5}{1.4}+\dfrac{5}{4.7}+...+\dfrac{5}{100.103}\) E= \(\dfrac{1}{15}+\dfrac{1}{35}+...+\dfrac{1}{2499}\)
A=2.(1/1.3 + 1/3.5 + 1/5.7 +.......+1/99.101)
=2.(1/1 + 1/3 + 1/5 + 1/5 + 1/7 +...+1/99 + 1/101)
=2.(1-1/101)
=2.(101/101-1/101)
=2.100/101
200/101
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B=2.(1/1.3+1/3.5+1/3.1+....+1/99.101)
=2.(1/1+1/3+1/3+1/5+1/3+1/7+....+1/99+1/101)
=2.(1/1+1/101)
=2.(101/101+1/101)
=2.102/101
=204/101
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C=1/2+1/3+1/3+1/4+....+1/99+1/100
=1/2+1/100
=50/100+1/100
=51/100
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chứng minh rằng :
a) \(\frac{a}{n\left(n+a\right)}=\frac{1}{n}-\frac{1}{n+a}\) ( n , a ϵ N* )
b) áp dụng câu a tính ;
\(A=\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{99.100}\)
\(B=\frac{5}{1.4}+\frac{5}{4.7}+...+\frac{5}{100.103}\)
\(C=\frac{1}{15}+\frac{1}{35}+...+\frac{1}{2499}\)
a) \(\frac{1}{n}-\frac{1}{n+a}=\frac{\left(n+a\right)-n}{n\left(n+a\right)}=\frac{a}{a\left(n+a\right)}\) (đpcm)
b) \(A=\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{99}-\frac{1}{100}=\frac{1}{2}-\frac{1}{100}=\frac{49}{100}\)
\(B=\frac{5}{3}.\left(1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+...+\frac{1}{100}-\frac{1}{103}\right)=\frac{5}{3}.\left(1-\frac{1}{103}\right)=\frac{5}{3}.\frac{102}{103}=\frac{170}{103}\)
\(C=\frac{1}{3.5}+\frac{1}{5.7}+...+\frac{1}{49.51}=\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{49}-\frac{1}{51}=\frac{1}{3}-\frac{1}{51}=\frac{16}{51}\)
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tính và so sánh: A=1/2.3+1/3.4+...+1/99.100 ; B=5/1.4+5/4.7+...+5/100.103
Ta có : \(A=\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{99.100}\)
\(A=\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{99}-\frac{1}{100}\)
\(A=\frac{1}{2}+\left(-\frac{1}{3}+\frac{1}{3}\right)+\left(-\frac{1}{4}+\frac{1}{4}\right)+...+\left(-\frac{1}{99}+\frac{1}{99}\right)-\frac{1}{100}\)
\(A=\frac{1}{2}+0+0+..+0-\frac{1}{100}\)
\(A=\frac{1}{2}-\frac{1}{100}=\frac{49}{100}\)
\(B=\frac{5}{1.4}+\frac{5}{4.7}+..+\frac{5}{100.103}\)
\(B=1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+...+\frac{1}{100}-\frac{1}{103}\)
\(B=1+\left(-\frac{1}{4}+\frac{1}{4}\right)+\left(-\frac{1}{7}+\frac{1}{7}\right)+...+\left(-\frac{1}{100}+\frac{1}{100}\right)-\frac{1}{103}\)
\(B=1+0+0+...+0-\frac{1}{103}\)
\(B=1-\frac{1}{103}=\frac{102}{103}\)
So sánh : A < B vì 49/100 < 102/103 (49.103 < 102 . 100)
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