Bạn tham khảo nhé!

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 = (1² + 3² + 5² + … + 97² + 99²) + 2.(1 + 3 + 5 + … + 97 + 99).

Đặt B = 1² + 3² + 5² + … + 99²

=> B = (1² + 2² + 3² + 4² + … + 100²) – 2².(1² + 2² + 3² + 4² + … + 50²)

Tính dãy tổng quát C = 1² + 2² + 3² + … + n²

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.

  A= \(\dfrac{1}{1.3}\)+\(\dfrac{1}{3.5}\)+\(\dfrac{1}{5.7}\)+\(\dfrac{1}{7.9}\)+...+\(\dfrac{1}{97.99}\)

2A= 1 - \(\dfrac{1}{3}\)+\(\dfrac{1}{3}\) - \(\dfrac{1}{5}\)+\(\dfrac{1}{5}\) - \(\dfrac{1}{7}\)+\(\dfrac{1}{7}\) - \(\dfrac{1}{9}\)+...+\(\dfrac{1}{97}\)-\(\dfrac{1}{99}\)

2A= 1-\(\dfrac{1}{99}\)

2A= \(\dfrac{98}{99}\)

  A= \(\dfrac{98}{99}\) : 2

A=\(\dfrac{49}{99}\)

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.

\(A=\dfrac{1}{1.3}+\dfrac{1}{3.5}+\dfrac{1}{5.7}+\dfrac{1}{7.9}+...+\dfrac{1}{97.99}\)
\(=\dfrac{1}{2}\left(\dfrac{2}{1.3}+\dfrac{2}{3.5}+\dfrac{2}{5.7}+\dfrac{2}{7.9}+...+\dfrac{2}{97.99}\right)\)
\(=\dfrac{1}{2}\left(1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{9}+...+\dfrac{1}{97}-\dfrac{1}{99}\right)\)
\(=\dfrac{1}{2}\left(1-\dfrac{1}{99}\right)\)
\(=\dfrac{1}{2}\cdot\dfrac{98}{99}\)
\(=\dfrac{49}{99}\)

 \(A=\frac{2}{1.3}+\frac{2}{3.5}+...+\frac{2}{97.99}\)

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

\(A=\frac{1}{1}-\frac{1}{99}\)

\(A=\frac{98}{99}\)

ta có A=1-1/3+1/2-1/5+..................1/95-1/97+1/97-1/99

        A=1-1/99

        A=98/99

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

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

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

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

~ Hok tốt ~

\(\)

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

Kết quả là:

 161651

  Đpá số: 161651

161651

ban nhe

tk nha@@@@@@@@@@@@@@@@@@@@@@@@2

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