\(\frac{2}{3.5}+\frac{2}{5.7}+\frac{2}{7.9}+...+\frac{2}{97.99}\)

\(=\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+...+\frac{1}{97}-\frac{1}{99}\)

\(=\frac{1}{3}+\left(\frac{1}{5}-\frac{1}{5}\right)+\left(\frac{1}{7}-\frac{1}{7}\right)+...+\left(\frac{1}{97}-\frac{1}{97}\right)-\frac{1}{99}\)

\(=\frac{1}{3}-\frac{1}{99}=\frac{32}{99}\)

~ Hok tốt ~

\(\)

Viết thành 2 . (1/3.5 + 1/5.7 + 1/7.9 + ...+ 1/97.99

Đặt A=\(\dfrac{2}{3.5}.\dfrac{2}{7.9}.....\dfrac{2}{99.101}\)

A=\(\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+...+\dfrac{1}{99}-\dfrac{1}{101}\)

A=\(\dfrac{1}{3}-\dfrac{1}{101}=\dfrac{98}{303}\)

Ta có: \(P=\dfrac{2}{3\cdot5}+\dfrac{2}{5\cdot7}+\dfrac{2}{7\cdot9}+\dfrac{2}{9\cdot11}+\dfrac{2}{11\cdot13}+\dfrac{2}{13\cdot15}\)

\(=\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+...+\dfrac{1}{13}-\dfrac{1}{15}\)

\(=\dfrac{1}{3}-\dfrac{1}{15}\)

\(=\dfrac{4}{15}\)

a.  

\(M=1.\left[\frac{1}{3}-\frac{1}{5}+.....\frac{1}{97}-\frac{1}{99}\right]\)

\(M=\frac{1}{3}-\frac{1}{99}=\frac{32}{99}\)

b.

\(N=\frac{3}{2}.\left[\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+...+\frac{1}{197}-\frac{1}{199}\right]\)

\(N=\frac{3}{2}.\left[\frac{1}{5}-\frac{1}{199}\right]=\frac{291}{995}\)

mk đầu tiên nha bạn

= \(\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{9}+...+\dfrac{1}{97}-\dfrac{1}{99}\)

\(=\dfrac{1}{3}-\dfrac{1}{99}=\dfrac{32}{99}\)

bạn làm thiếu rồi

phải chia 2 nữa

Khoảng cách giữa hai thừa số trong mỗi số hạng là 2, nhân 2 vế của A với 3 lần khoảng cách này ta được :

6A=1.3.6 + 3.5.6 + 5.7.6 + ... + 97.99.6

=1.3(5+1) + 3.5(7-1) + 5.7(9-3) + ... + 97.99(101-95)

=1.3.5 + 1.3 + 3.5.7 - 1.3.5 + 5.7.9 - 3.5.7 + ... + 97.99.101 - 95.97.99

=1.3.5 + 3 + 3.5.7 - 1.3.5 + 5.7.9 - 3.5.7+ ... + 97.99.101 - 97.97.99

=3+97.99.101

\(\frac{1+97.33.101}{1}=161651\)

Ta có :

B = 1.3 + 3.5 + 5.7 + 7.9 + ... + 97.99

6.B = 1.3.6 + 3.5.6 + 5.7.6 +...+ 97.99.6

6.B = 1.3.[ 5 - (-1) ] + 3.5.( 7 - 1 ) + 5.7.( 9 - 3 ) + ...+ 97.99.( 101 - 95 )

6.B = 1.3.5 - ( -1).3.5 + 3.5.7 - 1.3.5 + 5.7.9 - 3.5.7 + ... + 97.99.101 - 95.97.99

6.B = 97.99.101 - ( -1 ) .3.5

6.B = 97.99.101 + 1.3.5

6.B = 969918

=> B = 161653.

Bài 1:

a: \(\frac{2}{3\cdot5}+\frac{2}{5\cdot7}+\cdots+\frac{2}{97\cdot99}\)

\(=\frac13-\frac15+\frac15-\frac17+\cdots+\frac{1}{97}-\frac{1}{99}\)

\(=\frac13-\frac{1}{99}=\frac{32}{99}\)

b: \(\frac{1}{3\cdot5}+\frac{1}{5\cdot7}+\cdots+\frac{1}{97\cdot99}\)

\(=\frac12\left(\frac{2}{3\cdot5}+\frac{2}{5\cdot7}+\cdots+\frac{2}{97\cdot99}\right)\)

\(=\frac12\left(\frac13-\frac15+\frac15-\frac17+\cdots+\frac{1}{97}-\frac{1}{99}\right)\)

\(=\frac12\left(\frac13-\frac{1}{99}\right)=\frac12\cdot\frac{32}{99}=\frac{16}{99}\)

c: \(\frac{1}{18}+\frac{1}{54}+\frac{1}{108}+\cdots+\frac{1}{990}\)

\(=\frac{1}{3\cdot6}+\frac{1}{6\cdot9}+\frac{1}{9\cdot12}+\cdots+\frac{1}{30\cdot33}\)

\(=\frac13\left(\frac{3}{3\cdot6}+\frac{3}{6\cdot9}+\cdots+\frac{3}{30\cdot33}\right)\)

\(=\frac13\left(\frac13-\frac16+\frac16-\frac19+\cdots+\frac{1}{30}-\frac{1}{33}\right)\)

\(=\frac13\left(\frac13-\frac{1}{33}\right)=\frac13\cdot\frac{10}{33}=\frac{10}{99}\)

Bài 2:

Sửa đề: \(\frac{1}{41}+\frac{1}{42}+\cdots+\frac{1}{80}>\frac{7}{12}\)

Đặt \(A=\frac{1}{41}+\frac{1}{42}+\cdots+\frac{1}{80}\)

Ta có: \(\frac{1}{41}>\frac{1}{60}\)

\(\frac{1}{42}>\frac{1}{60}\)

...

\(\frac{1}{59}>\frac{1}{60}\)

\(\frac{1}{60}=\frac{1}{60}\)

DO đó: \(\frac{1}{41}+\frac{1}{42}+\cdots+\frac{1}{59}+\frac{1}{60}>\frac{1}{60}+\frac{1}{60}+\cdots+\frac{1}{60}+\frac{1}{60}=\frac{20}{60}=\frac13\) (1)

Ta có: \(\frac{1}{61}>\frac{1}{80}\)

\(\frac{1}{62}>\frac{1}{80}\)

...

\(\frac{1}{79}>\frac{1}{80}\)

\(\frac{1}{80}=\frac{1}{80}\)

Do đó: \(\frac{1}{61}+\frac{1}{62}+\cdots+\frac{1}{80}>\frac{1}{80}+\frac{1}{80}+\cdots+\frac{1}{80}=\frac{20}{80}=\frac14\) (2)

Từ (1),(2) suy ra \(\frac{1}{41}+\frac{1}{42}+\cdots+\frac{1}{80}>\frac13+\frac14\)

=>\(A>\frac13+\frac14\)

=>A>7/12

Ta có : S = 1.3 + 3.5 + 5.7 + .... + 97.99 + 99.101

=> 6S = 1.3.6 + 3.5.6 + 5.7.6 +...+ 97.99.6 + 99.101.6

           = 1.3.(5 + 1) + 3.5.(7 - 1) + 5.7.(9 - 3) + .... + 97.99.(101 - 95) + 99.101.(103 - 97)

           = 3 + 1.3.5 +  3.5.7 - 1.3.5 + 5.7.9 - 3.5.7 + ... + 97.99.101 - 95.97.99 + 99.101.103 - 97.99.101

           = 3 + 99.101.103

           =  1029900

=> 6S = 1029900

=> S = 171650

Ta có: A = 1.3 + 3.5 + 5.7 +…+ 97.99 + 99.101

A = 1.(1 + 2) + 3.(3 + 2) + 5.(5 + 2) + … + 97.(97 + 2) + 99.(99 + 2)

A = (1^2 + 3^2 + 5^2 + … + 97^2 + 99^2) + 2.(1 + 3 + 5 + … + 97 + 99).

Đặt B = 1^2 + 3^2 + 5^2 + … + 99^2

=> B = (1^2 + 2^2 + 3^2 + 4^2 + … + 100^2) – 2^2.(1^2 + 2^2 + 3^2 + 4^2 + … + 50^2)

Tính dãy tổng quát C = 1^2 + 2^2 + 3^2 + … + n^2

C = 1.(0 + 1) + 2.(1 + 1) + 3.(2 + 1) + … + n.[(n – 1) + 1]

C = [1.2 + 2.3 + … + (n – 1).n] + (1 + 2 + 3 + … + n)

C =  = n.(n + 1).[(n – 1) : 3 + 1 : 2] = n.(n + 1).(2n + 1) : 6

Áp dụng vào B ta được:

B = 100.101.201 : 6 – 4.50.51.101 : 6  = 166650

=> A = 166650 + 2.(1 + 99).50 : 2

=> A = 166650 + 5000 = 172650.

Đ/s: A = 172650.