A = 1/1.3 - 1/2.4 + 1/3.5 - 1/4.6 + ....+ 1/97.99 - 1/98.100

A = (1/1.3 + 1/3.5 + ...+ 1/97.99) - (1/2.4 - 1/4.6 + ...+ 1/98.100)

Đặt B = 1/1.3 + 1/3.5 + ..+ 1/97.99

C = 1/2.4 + 1/4.6 + ...+ 1/98.100

2B = 2/1.3 + 2/3.5 + ...+ 2/97.99

2B = 1/1 - 1/3 + 1/3- 1/5 + .. + 1/97 - 1/99

2B = 1/1 - 1/99

2B = 98/99

B = 98/99 : 2

B = 49/99

C = 1/2.4 + 1/4.6 + ... + 1/98.100

2C = 2/2.4 + 2/4.6 + ... + 2/98.100

2C = 1/2 - 1/4 + 1/4 - 1/6 + ... + 1/98 - 1/100

2C = 1/2 - 1/100

2C = 49/100

C = 49/100 : 2

C = 49/200

A = B - C

A = 49/99 - 49/200

B = 9800/19800 - 4851/19800

B = 4949/19800

E = 1.3 + 2.4 + 3.5 +...+ 97.99 + 98.100

A = 1.3 + 3.5 + 5.7 + ...+ 97.99

B = 2.4 + 4.6 + 6.8 + ... + 98.100

A = 1.3 + 3.5 + 5.7 + ... + 97.99

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

1.3.6 = 1.3.(5+ 1) = 1.3.5 + 1.3.1

3.5.6 = 3.5(7 - 1) = 3.5.7 - 1.3.5

5.7.6 = 5.7.(9 - 3) = 5.7.9 - 3.5.7

7.9.6 = 7.9.(11 - 5) = 7.9.11 - 5.7.9

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

97.99.6 = 97.99.(101 - 95) = 97.99.101 - 95.97.99

Cộng vế với vế ta có:

6A = 1.3.1 + 97.99.101

6A = 3 + 969903

6A = 969906

A = 969906 : 6

A = 161651

B = 2.4 + 4.6 + 6.8 + ... + 98.100

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

2.4.6 = 2.4.6

4.6.6 = 4.6.(8 - 2) = 4.6.8 - 2.4.6

6.8.6 = 6.8.(10 - 4) = 6.8.10 - 4.6.8

8.10.6 = 8.10.(12 - 6) = 8.10.12 - 6.8.10

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

98.100.6 = 98.100.(102 - 96) = 98.100.102 - 96.98.100

6B = 98.100.102

B = 98.100.102 : 6

B = 166600

E = A + B

E = 161651 + 166600

E = 328251

A = 1×3+3×5+5×7+...+ 97×99+99×101

 6A= 1×3×6+3×5×6+5×7×6+...+97×99×6+99×101×6

6A= 1×3×(5+1)+3×5×(7-1)+5×7×(9-3)+...+97×99×(101-95)+99×101×(103-97)

6A = 1×3×5-1×3+3×5×7-1×3×5+5×7×9-3×5×7+7×9×11-5×7×9+,,,+97×99×101-95×97×99+99×101×103-97×99×101

6A= 1×3+99×101×103

6A= 1029900

A= 171650

171650

\(C=\dfrac{4}{1.3}.\dfrac{9}{2.4}.\dfrac{16}{3.5}.\dfrac{25}{4.6}....\dfrac{9801}{9800}=\)

\(=\dfrac{2^2.3^2.4^2.5^2.....99^2}{1.2.3^2.4^2.5^2....98^2.99.100}=\dfrac{2.99}{100}=\dfrac{198}{100}=1,98\)

Câu 1:

A = 2.4 + 4.6 + 6.8 +...+ 98.100 + 100.102

6A = 2.4.6 + 4.6.6 +..+98.100.6 + 100.102.6

6A = 2.4.6 + 4.6.(8-2) +...+100.102.(104 - 98)

6A = 2.4.6 + 4.6.8 - 2.4.6 + ...+ 100.102.104 - 98.100.102

6A = 100.102.104

A = 100.102.104 : 6

A = 10200.104 : 6

A = 1060800 : 6

A = 176800

Câu 2:

B = 1.3 + 3.5 + 5.7 + ...+ 97.99 + 99.101

6B = 1.3.6 + 3.5.6 + ...+ 99.101.6

6B = 1.3.(5+1) . 3.5.(7-1) + ..+99.101.(103-97)

6B = 1.3.5 + 1.3.1 +3.5.7- 1.3.5 +...+99.101.103 - 97.99.101

6B = 1.3.1 + 99.101.103

6B = 3 + 9999.103

6B = 3 + 1029897

6B = 1029900

B = 1029900 : 6

B = 171650

e)đặt A=2^2+4^2+6^2+...+98^2+100^2

=2.2+4.4+6.6+...+98.98+100.100

=2.(4-2)+4.(6-2)+6.(8-2)+...+98.(100-2)+100.(102-2)

=2.4-4+4.6-8+6.8-12+...+98.100-196+100.102-200

=(2.4+4.6+6.8+...+98.100+100.102)-(4+8+12+...+196+200)

Đặt B=2.4+4.6+6.8+...+98.100+100.102

      6B=2.4.6+4.6.6+...+98.100.6+100.102.6

          =2.4.6+4.6.(8-2)+...+98.100.(102-96)+100.102.(104-98)

           =2.4.6+4.6.8-2.4.6+...+98.100.102-96.98.100+100.102.104-98.100.102

          =(2.4.6-2.4 .6)+...+(98.100.102-98.100.102)+100.102.104

          =100.102.104

      B=100.102.104/6=100.17.104=176800

Đặt C=4+8+12+...+196+200   Có 50 số hạng         Công thức tính số các số hạng  (số cuối-số đầu):khoảng cách+1

        =(200+4).50/2=5100                                         Công thức tính tổng số các số hạng  (số cuối +số đầu ). số các số hạng :2

Ta có A=176800-5100=171700

f) làm tương tự,hơi dài nên đành làm vậy,xin lỗi nha,nếu mà khó quá kết bạn với tớ ,tớ giải cho nha

Gợi ý đặt A=..

                  =...

                  =...

      Đặt B=...

          6B=...

              =...

             =...

       Đặt C=...

               =...

Ta có

\(\frac{1}{3.1}-\frac{1}{2.4}+\frac{1}{3.5}-\frac{1}{4.6}+...+\frac{1}{97.99}-\frac{1}{98.100}\)

= \(\left(\frac{1}{1.3}+\frac{1}{3.5}+...+\frac{1}{97.99}\right)-\left(\frac{1}{2.4}+\frac{1}{4.6}+...+\frac{1}{98.100}\right)\)

= \(\frac{1}{2}\left(1-\frac{1}{3}+...+\frac{1}{97}-\frac{1}{99}\right)-\frac{1}{2}\left(\frac{1}{2}-\frac{1}{4}+...+\frac{1}{98}-\frac{1}{100}\right)\)

= \(\frac{1}{2}\left(1-\frac{1}{99}\right)-\frac{1}{2}\left(\frac{1}{2}-\frac{1}{100}\right)=\frac{1}{2}.\frac{98}{99}-\frac{1}{2}.\frac{49}{100}\)

= \(\frac{49}{99}-\frac{49}{200}\)

= \(\frac{4949}{19800}\)

bn zô xem nha, ko hiểu thì cứ hỏi bn ấy nhá

http://olm.vn/hoi-dap/question/154321.html

Ta có: \(A=1\cdot99+2\cdot98+3\cdot97+\cdots+98\cdot2+99\cdot1\)

\(=2\left(1\cdot99+2\cdot98+\cdots+49\cdot51\right)+50\cdot50\)

\(=2\left\lbrack1\left(100-1\right)+2\left(100-2\right)+\cdots+49\left(100-49\right)\right\rbrack+2500\)

\(=2\cdot\left\lbrack100\left(1+2+\cdots+49\right)-\left(1^2+2^2+\cdots+49^2\right)\right\rbrack+2500\)

\(=2\cdot\left\lbrack100\cdot\frac{49\cdot50}{2}-\frac{49\cdot\left(49+1\right)\left(2\cdot49+1\right)}{6}\right\rbrack+2500\)

\(=2\left\lbrack50\cdot49\cdot50-\frac{49\cdot50\cdot99}{6}\right\rbrack+2500\)

\(=2\cdot\left\lbrack49\cdot50\cdot50-49\cdot25\cdot33\right\rbrack+2500\)

\(=2\cdot49\cdot25\cdot\left(2\cdot50-33\right)+2500\)

\(=49\cdot50\cdot67+2500=166650\)

Ta có: \(B=1\cdot2\cdot3+2\cdot3\cdot4+\ldots+17\cdot18\cdot19\)

\(=2\left(2-1\right)\left(2+1\right)+3\left(3-1\right)\left(3+1\right)+\cdots+18\left(18-1\right)\left(18+1\right)\)

\(=2\cdot\left(2^2-1\right)+3\left(3^2-1\right)+\cdots+18\left(18^2-1\right)\)

\(=\left(2^3+3^3+\cdots+18^3\right)-\left(2+3+\cdots+18\right)\)

\(=\left(1^3+2^3+\cdots+18^3\right)-\left(1+2+3+\cdots+18\right)\)

\(=\left(1+2+\cdots+18\right)^2-\left(1+2+\cdots+18\right)\)

\(=\left(18\cdot\frac{19}{2}\right)^2-18\cdot\frac{19}{2}=\left(9\cdot19\right)^2-9\cdot19=29070\)

Ta có: \(C=1\cdot4+2\cdot5+\cdots+100\cdot103\)

\(=1\left(1+3\right)+2\left(2+3\right)+\cdots+100\cdot\left(100+3\right)\)

\(=\left(1^2+2^2+\cdots+100^2\right)+3\left(1+2+\cdots+100\right)\)

\(=\frac{100\left(100+1\right)\left(2\cdot100+1\right)}{6}+\frac{3\cdot100\cdot101}{2}\)

\(=\frac{100\cdot101\cdot201}{6}+\frac{3\cdot100\cdot101}{2}=50\cdot101\cdot67+3\cdot50\cdot101\)

\(=50\cdot101\cdot70=3500\cdot101=353500\)

Ta có: \(D=1\cdot3+2\cdot4+3\cdot5+\cdots+97\cdot99+98\cdot100\)

\(=1\left(1+2\right)+2\left(2+2\right)+3\left(3+2\right)+\cdots+97\cdot\left(97+2\right)+98\cdot\left(98+2\right)\)

\(=\left(1^2+2^2+\cdots+98^2\right)+2\cdot\left(1+2+3+\cdots+98\right)\)

\(=\frac{98\cdot\left(98+1\right)\left(2\cdot98+1\right)}{6}+2\cdot\frac{98\cdot99}{2}\)

\(=\frac{98\cdot99\cdot197}{6}+98\cdot99=49\cdot33\cdot197+98\cdot99=49\cdot33\left(197+2\cdot3\right)\)

\(=49\cdot33\cdot203=328251\)

A=1-1/100