Đặt \(A=\) \(\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{a\left(a+1\right)}=\frac{49}{100}\)

\(\Rightarrow A=\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{a}-\frac{1}{a+1}=\frac{49}{100}\)

\(\Rightarrow A=\frac{1}{2}-\frac{1}{a+1}=\frac{49}{100}\)

\(\)\(\Rightarrow\frac{1}{a+1}=\frac{1}{2}-\frac{49}{100}\)

\(\)\(\Rightarrow\frac{1}{a+1}=\frac{1}{100}\Rightarrow a+1=100\Rightarrow a=100-1\)

\(\Rightarrow a=99\)

                  Vậy \(a=99\)k cho mik nha :))

bài này dài lắm nhưng kết quả là 99 

mình ngại giải dài lắm !!!

\(\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{a.\left(a+1\right)}=\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{a}-\frac{1}{a+1}=\frac{1}{2}-\frac{1}{a+1}=\frac{49}{100}\)

\(\Rightarrow\frac{1}{a+1}=\frac{1}{100}\Rightarrow a+1=99\)

99

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8

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6

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het

\(\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{a.\left(a+1\right)}=\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{a}-\frac{1}{a+1}=\frac{1}{2}-\frac{1}{a+1}=\frac{49}{100}\)

\(\Rightarrow\frac{1}{a+1}=\frac{1}{100}\Rightarrow a+1=100\Rightarrow a=99\)

dung roi 

chỉ bt lm phần cuối thôi

D = 1.2.3 + 2.3.4 + ... + 98.99.100

4D = 1.2.3.4 + 2.3.4.4 + ... + 98.99.100.4

4D = 1.2.3.4 + 2.3.4.(5 - 1) + ... + 98.99.100.(101 - 97)

4D = 1.2.3.4 + 2.3.4.5 - 1.2.3.4 + ... + 98.99.100.101 - 97.98.99.100

4D = 98.99.100

D = 98.99.100 : 4

D = 242550

a, 6A = 1.3.6 + 3.5 .6 + 5.7.6 + ... + 49.51.6

        = 1.3.6 + 3.5.(7-1) + 5.7.(9-3) + ... + 49.51.(53-47)

        = 18 + 3.5.7 - 1.3.5 + 5.7.9 - 3.5.7 + ... + 49.51.53 - 47.49.51

       = 18 - 1.3.5 + 49.51.53

       = 132450

=> A = 132450 : 6 = 22075

b , Giải giống câu a

 6B = 2.4.6 + 4.6.6 + 6.8.6 + ... + 98.100.6

      = 2.4.6 + 4.6.(8-2) + 6.8.(10-4) + ... + 98.100.(102-96)

  Đến đây làm giống câu a  : Phá ngoặc rồi triệt tiêu đi!
c , 9C= 1.4.9 + 4.7.9 + 7.10.9 + ... + 301.304.9

        = 1.4.9 + 4.7.(10 - 1)  + 7.10.( 13-4) + ...  + 301.304.(307-298)

   Đến đây làm giống câu a  : Phá ngoặc rồi triệt tiêu đi!

CÒn câu d thì Nguyễn Phương Uyên làm đúng rồi

CHÚ Ý : Cách giải chung: 

 Nhân biểu thức cần tính với 3 lần khoảng cách giữa các số  để xuất hiện các số hạng đối nhau !

Ai trả lời nhanh mình tích cho nhé!

\(A=\frac{1}{1.2.3}+\frac{1}{2.3.4}+...+\frac{1}{98.99.100}\)

\(A=\frac{1}{2}.\left(\frac{2}{1.2.3}+\frac{2}{2.3.4}+...+\frac{2}{98.99.100}\right)\)

\(A=\frac{1}{2}.\left(\frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{2.3}-\frac{1}{3.4}+...+\frac{1}{98.99}-\frac{1}{99.100}\right)\)

\(A=\frac{1}{2}.\left(\frac{1}{2}-\frac{1}{9900}\right)\)

\(A=\frac{1}{2}.\frac{4949}{9900}\)

\(A=\frac{4949}{19800}\)

A = 1 . 3 + 3 . 5 + 5 . 7 + ... + 49 . 51

=  1 . 51

= 51

B = 2 . 4 + 4 . 6 + 6 . 8 + ... + 98 . 100

= 2 . 100

= 200

C = 1 . 4 + 4 . 7 + 7 . 10 + ... + 301 . 304

= 1 . 304

= 304

D = 1 + 1 . 1! + 2 . 2! + 3 . 3!  + ... + 100 . 100!

= 1 . 100

= 100

E = 2² + 4² + ... + ( 2n )²

= 2² . ( 2n )²

= 2n⁴ 

bn chưa ghi rõ đề bài , theo mik nghĩ bn cần tìm bao nhiều số chứ koo pk tính tổng 

A = 1 .51

A = 51

B= 2.100

B= 200

C= 1.304 

C= 304

D= 1.100!

D= 100!

E=2² + 4² + .....+ ( 2n) ²

E = 2² . 2n²

E= 2n⁴

hok tốt !

a, \(B=\dfrac{1}{3}+\dfrac{1}{3^2}+\dfrac{1}{3^3}+...+\dfrac{1}{3^{2004}}+\dfrac{1}{3^{2005}}\\ 3B=3+\dfrac{1}{3}+\dfrac{1}{3^2}+...+\dfrac{1}{3^{2003}}+\dfrac{1}{3^{2004}}\\ 3B-B=\left(3+\dfrac{1}{3}+\dfrac{1}{3^2}+\dfrac{1}{3^3}+...+\dfrac{1}{3^{2003}}+\dfrac{1}{3^{2004}}\right)-\left(\dfrac{1}{3}+\dfrac{1}{3^2}+\dfrac{1}{3^3}+...+\dfrac{1}{3^{2004}}+\dfrac{1}{3^{2005}}\right)\\2B=3-\dfrac{1}{3^{2005}}\\ B=\dfrac{3-\dfrac{1}{3^{2005}}}{2}\)

b,

\(A=1+5+5^2+5^3+...+5^{49}+5^{50}\\ 5A=5+5^2+5^3+5^4+...+5^{50}+5^{51}\\ 5A-A=\left(5+5^2+5^3+5^4+...+5^{50}+5^{51}\right)-\left(1+5+5^2+5^3+...+5^{49}+5^{50}\right)\\ 4A=5^{51}-1\\ A=\dfrac{5^{51}-1}{4}\)

c,

\(A=\left(\dfrac{1}{2^2}-1\right)\left(\dfrac{1}{3^2}-1\right)\left(\dfrac{1}{4^2-1}\right)......\left(\dfrac{1}{100^2-1}\right)\\ A=\left(\dfrac{1}{4}-1\right)\left(\dfrac{1}{9}-1\right)\left(\dfrac{1}{16}-1\right)......\left(\dfrac{1}{10000}-1\right)\\ A=\dfrac{3}{4}\cdot\dfrac{8}{9}\cdot\dfrac{15}{16}\cdot\cdot\cdot\cdot\dfrac{9999}{10000}\\ A=\dfrac{1\cdot3}{2\cdot2}\cdot\dfrac{2\cdot4}{3\cdot3}\cdot\dfrac{3\cdot5}{4\cdot4}\cdot\cdot\cdot\cdot\dfrac{99\cdot101}{100\cdot100}\\ A=\dfrac{1\cdot2\cdot3\cdot\cdot\cdot\cdot99}{2\cdot3\cdot4\cdot\cdot\cdot\cdot100}\cdot\dfrac{3\cdot4\cdot5\cdot\cdot\cdot\cdot101}{2\cdot3\cdot4\cdot\cdot\cdot\cdot100}\\ A=\dfrac{1}{100}\cdot\dfrac{101}{2}\\ A=\dfrac{101}{200}\)

d,

\(A=2^{100}-2^{99}+2^{98}-2^{97}+...+2^2-2\\ A=\left(2^{100}+2^{98}+...+2^2\right)-\left(2^{99}+2^{97}+...+2^1\right)\)

Đặt \(A=B-C\)

\(\Rightarrow B=\left(2^{100}+2^{98}+...+2^2\right)vàC=\left(2^{99}+2^{97}+...+2^1\right)\)

\(B=2^{100}+2^{98}+...+2^2\\ 4B=2^{102}+2^{100}+...+2^4\\ 4B-B=\left(2^{102}+2^{100}+...+2^4\right)-\left(2^{100}+2^{98}+...+2^2\right)\\ 3B=2^{102}-2^2\\ B=\dfrac{2^{102}-2^2}{3}\left(1\right)\)

\(C=2^{99}+2^{97}+...+2^1\\ 4C=2^{101}+2^{99}+...+2^3\\ 4C-C=\left(2^{101}+2^{99}+...+2^3\right)-\left(2^{99}+2^{97}+...+2\right)\\ 3C=2^{101}-2\\ C=\dfrac{2^{101}-2}{3}\left(2\right)\)

Từ (1) và (2) ta có :

\(A=\dfrac{2^{102}-2^2}{3}-\dfrac{2^{101}-2}{3}\\ A=\dfrac{2^{102}-2^2-2^{101}+2}{3}\\ A=\dfrac{2^{102}-2^{101}+2}{3}\)