\(C=\frac{2}{3}.\left(\frac{3}{1.4}+\frac{3}{4.7}+........+\frac{3}{61.64}\right)\)

\(=\frac{2}{3}.\left(1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+.........+\frac{1}{61}-\frac{1}{64}\right)\)

\(=\frac{2}{3}.\left(1-\frac{1}{64}\right)\)

\(=\frac{2}{3}.\frac{63}{64}\)

\(=\frac{21}{32}\)

\(C=\frac{2}{1.4}+\frac{2}{4.7}+\frac{2}{7.10}+...+\frac{2}{61.61}\)

\(=2.\frac{1}{3}.\left(\frac{3}{1.4}+\frac{3}{4.7}+\frac{3}{7.10}+...+\frac{3}{61.64}\right)\)

\(=\frac{2}{3}.\left(\frac{1}{1}-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{10}+...+\frac{1}{61}-\frac{1}{64}\right)\)

\(=\frac{2}{3}.\left(1-\frac{1}{64}\right)\)

\(=\frac{2}{3}.\frac{63}{64}\)

\(=\frac{21}{32}\)

\(C=\frac{2}{1.4}+\frac{2}{4.7}+.....+\frac{2}{61.64}\)

\(\Rightarrow\frac{3}{2}C=1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+....+\frac{1}{61}-\frac{1}{64}\)

\(\Rightarrow C=\left(1-\frac{1}{64}\right):\frac{3}{2}=\frac{21}{32}\)

S = 1.4 + 4.7 + 7.10 + 10.13 + ... + 61.64

1.4.9 = 1.4.(7 + 2) = 1.4.7 + 1.4.2

4.7.9 = 4.7.(10 - 1) = 4.7.10 - 1.4.7

7.10.9 = 7.10.(13 - 4) = 7.10.13 - 4.7.10

10.13.9 = 10.13.(16 - 7) = 10.13.16 - 7.10.13

.......................................................................

61.64.9 = 61.64.(67 - 58) = 61.64.67 - 58.61.64

Cộng vế với vế ta có:

1.4.9 + 4.7.9 + 7.10.9 +...+ 61.64.9 = 1.4.2 + 61.64.67

9(1.4 + 4.7 + 7.10+ ...+ 61.64) = 261576

  1.4 + 4.7 + 7.10 +...+ 61.64 = 261576 : 9

1.4 + 4.7 + 7.10 + ... + 61.64 = 29064 

A = \(\dfrac{15}{1.4}\) + \(\dfrac{15}{4.7}\) + \(\dfrac{15}{7.10}\) + ... + \(\dfrac{15}{61.64}\)

A = \(5.\left(\dfrac{3}{1.4}+\dfrac{3}{4.7}+\dfrac{3}{7.10}...+\dfrac{3}{61.64}\right)\)

A = 5.( \(\dfrac{1}{1}\) - \(\dfrac{1}{4}\) + \(\dfrac{1}{4}\) - \(\dfrac{1}{7}\) + \(\dfrac{1}{7}\) - \(\dfrac{1}{10}\) + ... + \(\dfrac{1}{61}\) - \(\dfrac{1}{64}\))

A = 5.( \(\dfrac{1}{1}\) - \(\dfrac{1}{64}\))

A = 5. \(\dfrac{63}{64}\)

A = \(\dfrac{315}{64}\)

\(A=\dfrac{3}{1\cdot4}+\dfrac{3}{4\cdot7}+\dfrac{3}{7\cdot10}+...+\dfrac{3}{61\cdot64}+\dfrac{3}{64\cdot67}\)

\(A=1-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{10}+...+\dfrac{1}{61}-\dfrac{1}{64}+\dfrac{1}{64}-\dfrac{1}{67}\)

\(A=1-\dfrac{1}{67}\) < 1

=> A<1

Ta có:

\(A=\dfrac{3}{1.4}+\dfrac{3}{4.7}+\dfrac{3}{7.10}+...+\dfrac{3}{61.64}+\dfrac{3}{64.67}\)

\(=3.\dfrac{1}{3}.\left(1-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{10}+...+\dfrac{1}{61}-\dfrac{1}{64}+\dfrac{1}{64}-\dfrac{1}{67}\right)\)

\(=3.\left(1-\dfrac{1}{67}\right)\)

\(=3.\dfrac{66}{67}\)

\(=\dfrac{198}{67}\)

Vì \(\dfrac{198}{67}\) có tử lớn hơn mẫu nên \(\dfrac{198}{67}>1\)

Vậy \(A>1\)

sửa bài:

...  \(=1-\dfrac{1}{67}\)

\(=\dfrac{66}{67}\)

Vì \(\dfrac{66}{67}\) có tử nhỏ hơn mẫu nên \(\dfrac{66}{67}< 1\)

Vậy \(A< 1\)

\(A=\frac{2}{1.4}+\frac{2}{4.7}+\frac{2}{7.10}+...+\frac{2}{97.100}\)

\(A=2\left(\frac{1}{1.4}+\frac{1}{4.7}+\frac{1}{7.10}+...+\frac{1}{97.100}\right)\)

\(A=2\left(1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{10}+...+\frac{1}{97}-\frac{1}{100}\right)\)

\(A=2\left(1-\frac{1}{100}\right)\)

\(A=2.\frac{99}{100}=..............\)

Tự làm nốt nha

what?

\(=-\left(\dfrac{1}{1}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{7}+...+\dfrac{1}{61}-\dfrac{1}{64}\right)=-\dfrac{1}{63}\)

Bài 1 :

\(S=1.3+3.5+5.7+...+99.101=3+15+35+...9999\)

Ta thấy :

\(3=2^2-1\)

\(15=4^2-1\)

\(35=6^2-1\)

.....

\(9999=100^2-1\)

\(\Rightarrow S=2^2+4^2+...+100^2-\left(1\right).\left(\left(100-2\right):2+1\right)\)

\(\Rightarrow S=\dfrac{100.\left(100+1\right)\left(2.100+1\right)}{6}-51\)

\(\Rightarrow S=\dfrac{100.101.201}{6}-51=338299\)

nhanh len nhé mik đang cần gấp ai lam trước mik tích cho

Bài 6 :

\(C=1^2+2^2+...+100^2=\dfrac{100.\left(100+1\right)\left(2.100+1\right)}{6}=\dfrac{100.101.201}{6}=338350\)

Bài 9 :

\(S=1^2+2^2+3^2+...+99^2=\dfrac{99.\left(99+1\right)\left(2.99+1\right)}{6}=\dfrac{99.100.199}{6}=328350\)