D = 2.4 + 3.5 + 4.6 + 5.7 + 6.8 + 7.9 + ...+ 97.99 + 98.100

D = (2.4 + 4.6 +...+ 98.100) + (3.5 + 5.7 + 7.9 + ..+ 97.99)

Đặt A = 2.4 + 4.6 +...+ 98.100

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

Khi đó: D = A + B

Xét tổng A : A = 2.4 + 4.6 + ..+ 98.100

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

6A = 2.4.6 + 4.6.(8 - 2) +...+ 98.100.(102-96)

6A = 2.4.6 + 4.6.8 - 2.4.6 + ...+ 98.100.102 - 96.98.100

6A = 98.100.102

A = 98.100.102 : 6

A = 166600

Xét tổng B = 3.5 + 5.7 + ...+ 97.99

6B = 3.5.6 + 5.7.6 + ..+ 97.99.6

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

6B = 3.5.7 - 1.3.5 + 5.7.9 - 3.5.7 + ... + 97.99.101 - 97.99.95

6B = - 1.3.5 + 97.99.101

6B = - 15 + 969903

6B = 969888

B = 969888 : 6

B = 161648

D = A + B

D = 166600 + 161648

D = 328248

\(S=\frac{1.3}{3.5}+\frac{2.4}{5.7}+\frac{3.5}{7.9}+...+\frac{\left(n-1\right)\left(n+1\right)}{\left(2n-1\right)\left(2n+1\right)}+...+\frac{1002.1004}{2005.2007}\)

\(\Rightarrow S=\frac{\left(2-1\right)\left(2+1\right)}{\left(2.2-1\right)\left(2.2+1\right)}+\frac{\left(3-1\right)\left(3+1\right)}{\left(3.2-1\right)\left(3.2+1\right)}+...+\frac{\left(n-1\right)\left(n+1\right)}{\left(2n-1\right)\left(2n+1\right)}\)

\(+..+\frac{\left(1003-1\right)\left(1003+1\right)}{\left(1003.2-1\right)\left(1003.2+1\right)}\)

\(\Rightarrow S=\frac{1}{4}-\frac{3}{8}\left(\frac{1}{2.2-1}-\frac{1}{2.2+1}\right)+\frac{1}{4}-\frac{3}{8}\left(\frac{1}{3.2-1}-\frac{1}{3.2+1}\right)+...\)

\(+\frac{1}{4}-\frac{3}{8}\left(\frac{1}{2n-1}-\frac{1}{2n+1}\right)+...+\frac{1}{4}-\frac{3}{8}\left(\frac{1}{1003.2-1}-\frac{1}{1003.2+1}\right)\)

\(\Rightarrow S=1002.\frac{1}{4}-1002.\frac{3}{8}\left(\frac{1}{2.2-1}-\frac{1}{2.2+1}+\frac{1}{3.2-1}-...-\frac{1}{1003.2+1}\right)\)

\(\Rightarrow S=\frac{501}{2}-\frac{1503}{4}\left(\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{2005}-\frac{1}{2007}\right)\)

\(\Rightarrow S=\frac{501}{2}-\frac{1503}{4}\left(\frac{1}{3}-\frac{1}{2007}\right)\)

\(\Rightarrow S=\frac{501}{2}-\frac{1503}{4}.\frac{668}{2007}\)

\(\Rightarrow S=\frac{501}{2}-\frac{27889}{223}\)

\(\Rightarrow S=125,4372197\)

\(\)

thx  you

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

b: 6B=2*4*6+4*6*6+6*8*6+...+46*48*6+48*50*6

=2*4*6-2*4*6+4*6*8-4*6*8+...-44*46*48+46*48*50-46*48*50+48*50*52

=48*50*52

=>B=20800

d: 9D=1*4*9+4*7*9+...+46*49*9

=1*4*2+1*4*7-1*4*7+1*7*10-1*7*10+...+46*49*52-46*49*43

=1*2*4+46*49*52

=117216

=>D=13024

a: 

Đặt \(A=\frac{1}{1.3}+\frac{1}{2.4}+...+\frac{1}{8.10}\)

\(2A=\frac{2}{1.3}+\frac{2}{2.4}+...+\frac{2}{8.10}\)

\(2A=1-\frac{1}{3}+\frac{1}{2}-\frac{1}{4}+...+\frac{1}{8}-\frac{1}{10}\)

\(2A=1-\frac{1}{10}\)

\(2A=\frac{9}{10}\)

\(A=\frac{9}{10}:2=\frac{9}{20}\)

=\(\frac{1}{2}\left(\frac{2}{1.3}+...+\frac{2}{8.10}\right)\)

=\(\frac{1}{2}\left(\frac{1}{1}-\frac{1}{3}+\frac{1}{2}-\frac{1}{4}...+\frac{1}{8}-\frac{1}{10}\right)\)

( chắc chắn có số trái dấu ở phía sau, nên còn lại như sau)

=\(\frac{1}{2}\left(1-\frac{1}{10}\right)=\frac{1}{2}.\frac{9}{10}=\frac{9}{20}\)

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

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

A=\(\frac{1+97.33.101}{2}\) = 161 651 

161651