Đặt A = \(1.3.5+3.5.7+5.7.9+..+93.95.97+95.97.99\)  

\(8A=1.3.5.8+3.5.7.8+...+93.95.97.8+95.97.99.8\) 

\(8A=1.3.5.\left(1+7\right)+3.5.7.\left(9-1\right)+...+95.97.99\left(101-93\right)\) 

\(8A=1.3.5.7+15+3.5.7.9-1.3.5.7+...+95.97.99.101-93.95.97.99\) 

\(8A=15+95.97.99.101\) 

\(\Rightarrow\) \(A=\frac{15+95.97.99.101}{8}\) 

\(KL:........\) 

\(G=1.3.5+3.5.7+5.7.9+...+95.97.99\)

\(G=1+99.\left(3+5+7+...+97\right)\)\

\(G=100.\left[\left(3+97\right)+\left(5+95\right)+...+\left(49+51\right)\right]\)

\(G=100.\left(100.24\right)\)

\(G=100.2400=240000\)

240000

a) 9 + 99 + 999 + ... + 999999

= (10 - 1) + (100 - 1) + (1000 - 1) + ... + (1000000 - 1)

= (10¹ + 10² + 10³ + ... + 10⁶) - (1.6)

= 1111110 - 6 = 1111104

b) 1 + 11 + 111 + ... + 1111111

= 1 + (10¹ + 1) + (10² + 10¹ + 1) + ... + (10⁶ + 10⁵ + 10⁴ + 10³ + 10² + 10¹ + 1)

= 10¹ . 6 + 10² . 5 + 10³ . 4 + ... + 10⁶. 1) + (1 + 1.6)

= 60 + 500 + 4000 + ... + 1000000 + 7

= 1234560 + 7 = 1234567

c) C = 1.2 + 2.3 + 3.4 + 4.5 + ... + 98.99

3C = 1.2.3 + 2.3.3 + 3.4.3 + 4.5.3 + ... + 98.99.3

3C = 1.2.3 + 2.3.(4 - 1) + 3.4.(5 - 2) + ... + 98.99.(100 - 97)

3C = 1.2.3 + 2.3.4 - 2.3.1 + 3.4.5 - 3.4.2 +...+ 98.99.100 - 98.99.97

3C = 98.99.100

C = \(\dfrac{98.99.100}{3}\) = 323400

d) D = 1.3.5 + 3.5.7 + 5.7.9 + ... + 95.97.99

8D = 1.3.5.8 + 3.5.7.8 + 5.7.9.8 + ... + 95.97.99.8

8D = 1.3.5.(7 + 1) + 3.5.7.(9 - 1) + 5.7.9.(11 - 3) + ... + 95.97.99.(101 - 93)

8D = 1.3.5.7 + 1.3.5.1 + 3.5.7.9 - 3.5.7.1 + 5.7.9.11 - 5.7.9.3 + ... + 95.97.99.101 - 95.97.99.93

8D = 1.3.5.1 + 95.97.99.101

D = \(\dfrac{1.3.5.1+95.97.99.101}{8}=15517600\)

1.3.5.8 + 3.5.7.8 + 5.7.9.8 + … + 95.97.99.8

= 1.3.5(7 + 1) + 3.5.7(9 - 1) + 5.7.9(11 - 3) + … + 95.97.99(101 - 93)

= 1.3.5.7 + 15 + 3.5.7.9 - 1.3.5.7 + 5.7.9.11 - 3.5.7.9 + … + 95.97.99.101 - 93.95.97.99

= 15 + 95.97.99.101

=> \(A=\frac{15.95+97.99.101}{8}\)

Tách ra rồi tính

Bài này rất dễ may cho bạn là hôm nay minh rảnh nên mới giải cho tiết.Thôi tớ làm nhé!

Theo bài ra ta có :

A =1.3.5+3.5.7+5.7.9+...+97.99.101

8A=8.(1.3.5+3.5.7+5.7.9+...+97.99.101)

8A=1.3.5.8+3.5.7.8+5.7.9.8+...+97.99.101.8

8A=1.3.5.(7+1)+3.5.7(9-1)+5.7.9.(11-3)+...+97.99.101.(103-95)

8A=1.3.5.7+1.1.3.5+3.5.7.9-1.3.5.7+5.7.9.11-3.5.7.9+...+97.99.101.103-95.97.99.101

8A=1.1.3.5+97.99.101.103

8A=15+99900009

8A=99900024

A =99900024:8

A =12487503

Vậy A=12487503

A = 1/4.( 4/1.3.5 + 4/3.5.7+ ....+ 4/95.97.99)

    = 1/4 .( 1/ 1.3 - 1/3.5 + 1/3.5 - 1/5.7 + .......+ 1/95.97 - 1/97.99)

   = 1/4( 1/1.3 - 1/97.99)

    = 1/4 . 9499/29397

\(A=\frac{1}{1.3.5}+\frac{1}{3.5.7}+\frac{1}{5.7.9}+...+\frac{1}{95.97.99}\)

\(A=4.\left(\frac{1}{1.3}-\frac{1}{3.5}+\frac{1}{3.5}-\frac{1}{5.7}+\frac{1}{5.7}-\frac{1}{7.9}+...+\frac{1}{95.97}-\frac{1}{97.99}\right)\)

\(A=4.\left(\frac{1}{1.3}-\frac{1}{97.99}\right)\)

\(A=4.\frac{3200}{9603}=\frac{12800}{9603}\)

\(A=\frac{1}{1.3.5}+\frac{1}{3.5.7}+\frac{1}{5.7.9}+...+\frac{1}{95.97.99}\)

\(A=\frac{1}{4}.\left(\frac{1}{1.3.5}+\frac{1}{3.5.7}+\frac{1}{5.7.9}+...+\frac{1}{95.97.99}\right)\)

\(A=\frac{1}{4}.\left(\frac{1}{1.3}-\frac{1}{3.5}+\frac{1}{3.5}-\frac{1}{5.7}+\frac{1}{5.7}-\frac{1}{7.9}+...+\frac{1}{95.97}-\frac{1}{97.99}\right)\)

\(A=\frac{1}{4}.\left(\frac{1}{1.3}-\frac{1}{97.99}\right)\)

\(A=\frac{1}{4}.\frac{3200}{9603}\)

\(A=\frac{800}{9603}\)

Bài trc mik làm lộn :)))
~ Hok tốt ~
 

a)\(A=\frac{1}{5}+\frac{1}{5^2}+....+\frac{1}{5^2^5}\)    <=>\(5A=1+\frac{1}{5}+\frac{1}{5^2}+....+\frac{1}{5^{24}}\)

                                                             <=>\(5A-A=(1+\frac{1}{5}+...+\frac{1}{5^{24}})-(\frac{1}{5}+\frac{1}{5^2}+...+\frac{1}{5^{25}})\)

                                                             <=>\(4A=1-\frac{1}{5^{25}}\)  <=>\(A=\frac{(5^{25^{ }}-1)}{5^{25}}\div4\)

B = \(\frac{1}{1.3.5}+\frac{1}{3.5.7}+....+\frac{1}{95.97.99}\)

B = \(\frac{1}{4}.\left(\frac{5-1}{1.3.5}+\frac{7-3}{3.5.7}+...+\frac{99-95}{95.97.99}\right)\)

B = \(\frac{1}{4}.\left(\frac{5}{1.3.5}-\frac{1}{1.3.5}+\frac{7}{3.5.7}-\frac{3}{3.5.7}+...+\frac{99}{95.97.99}-\frac{95}{95.97.99}\right)\)

B = \(\frac{1}{4}.\left(\frac{1}{1.3}-\frac{1}{3.5}+\frac{1}{3.5}-\frac{1}{5.7}+...+\frac{1}{95.97}-\frac{1}{97.99}\right)\)

B = \(\frac{1}{4}.\left(\frac{1}{3}-\frac{1}{97.99}\right)\)

B = \(\frac{1}{4}.\frac{3200}{9603}\)

B = \(\frac{800}{9603}\)