Bài 1: Bạn ơi số 2004 không thuộc dãy A 

A có số số hạng là: (2005 - 5) : 4 + 1 = 501 (số hạng)

A = (2005 + 5) x 501 : 2 = 503505

Bài 2:

a) B = 4 . 1 + 4 . 5 + 4 . 5² + 4 . 5³ + ... + 4 . 5¹⁰⁰⁰

=> B = 4 . ( 1 + 5 + 5² + 5³ + .... + 5¹⁰⁰⁰)

b) 5 + 5² + 5³ + .... + 5¹⁰⁰⁰ có tận cùng là 0 (Do các lũy thừa với cơ số là 5 thì có tận cùng là 5 [25] mà ở đây có số số hạng là chẵn)

=> 1 + 5 + 5² + 5³ + .... + 5¹⁰⁰⁰ có tận cùng là số 1

=> 4 . ( 1 + 5 + 5² + 5³ + .... + 5¹⁰⁰⁰) có tận cùng là 4.

Vậy B có tận cùng là 4.

Bài 3:

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

=> A = 1.(1 + 2) + 3.(3 + 2) + 5.(5 + 2) + .... + 97.(97 + 2)

=> A = 1² + 1.2 + 3² + 3.2 + 5² + 5.2 + .... + 97² + 97.2

=> A = (1² + 3² + 5² + .... + 97²) + (1.2 + 3.2 + 5.2 + .... + 97.2)

=> A = (1² + 3² + 5² + .... + 97²) + 2(1 + 3 + 5 + .... + 97)

=> A =  (1² + 3² + 5² + .... + 97²) + 2 { (97 + 1) . [(97 - 1) : 2 + 1] : 2 }

=> A =  (1² + 3² + 5² + .... + 97²) + 24802

Đặt B =  (1² + 3² + 5² + .... + 97²)

=> B = 1.1 + 3.3 + 5.5 + .... + 97.97

=> B = 1.(0 + 1) + 3.(1 + 2) + 5.(4 + 1) + ..... + 97.(96 + 1)

=> B = 0 + 1.1 + 3 + 2.3 + 5 + 4.5 + .... + 97 + 96.97

=> B = (0 + 3 + 5 + .... + 97) + (1.1 + 2.3 + 4.5 + .... + 96.97)

=> B =  2400 + \(\frac{\left(97-1\right).97\left(97+1\right)}{6}\)

=> B = 2400 + 152096 = 154496

=> A = 154496 + 4802 = 159298

(Làm tương tự ở câu 2 nha)

bạn coi đề 1 sai rồi

\(a,\dfrac{3}{5}+\dfrac{3}{5\cdot9}+\dfrac{3}{9\cdot13}+....+\dfrac{3}{97\cdot101}\)

\(=\dfrac{3}{4}\cdot\left(\dfrac{4}{5}+\dfrac{4}{5\cdot9}+\dfrac{4}{9\cdot13}+....+\dfrac{4}{97\cdot101}\right)\)

\(=\dfrac{3}{4}\cdot\left(1-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{9}+\dfrac{1}{9}-\dfrac{1}{13}+....+\dfrac{1}{97}-\dfrac{1}{101}\right)\)

\(=\dfrac{3}{4}\cdot\left(1-\dfrac{1}{101}\right)\)

\(=\dfrac{3}{4}\cdot\dfrac{100}{101}\)

\(=\dfrac{75}{101}\)

\(b,\left(1+\dfrac{1}{2}\right)\cdot\left(1+\dfrac{1}{3}\right)\cdot\left(1+\dfrac{1}{4}\right)\cdot....\cdot\left(1+\dfrac{1}{99}\right)\)

\(=\dfrac{3}{2}\cdot\dfrac{4}{3}\cdot\dfrac{5}{4}\cdot....\cdot\dfrac{100}{99}\)

\(=\dfrac{100}{2}=50\)

Tính nhanh:

a) \(\dfrac{3}{5}+\dfrac{3}{5.9}+\dfrac{3}{9.13}+...+\dfrac{3}{97.101}\)

= \(\dfrac{3}{4}\left(1-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{9}+\dfrac{1}{9}-\dfrac{1}{13}+...+\dfrac{1}{97}-\dfrac{1}{101}\right)\)

= \(\dfrac{3}{4}\left(1-\dfrac{1}{101}\right)\)

= \(\dfrac{3}{4}\times\dfrac{100}{101}\)

= \(\dfrac{75}{101}\)

b) \(\left(1+\dfrac{1}{2}\right)\left(1+\dfrac{1}{3}\right)\left(1+\dfrac{1}{4}\right)...\left(\dfrac{1}{98}+1\right)\left(\dfrac{1}{99}+1\right)\)

\(=\dfrac{3}{2}.\dfrac{4}{3}.\dfrac{5}{4}...\dfrac{99}{98}.\dfrac{100}{99}\)

\(=\dfrac{3.4.5...99.100}{2.3.4...98.99}\)

\(=\dfrac{100}{2}\)

\(=50\)

Vì GTTĐ luôn lớn hơn hoặc bằng 0

=> \(\left|x+\frac{1}{1\cdot5}\right|+\left|x+\frac{1}{5\cdot9}\right|+...+\left|x+\frac{1}{397\cdot401}\right|=100x\ge0\)

=> \(x\ge0\)

=> \(x+\frac{1}{1\cdot5}+x+\frac{1}{5\cdot9}+...+x+\frac{1}{397\cdot401}=100x\)

=> \(\left(x+x+...+x\right)+\left(\frac{1}{1\cdot5}+\frac{1}{5\cdot9}+...+\frac{1}{397\cdot401}\right)=100x\)

Sau đấy tính vế phải, lấy 100x - vế trái x, rồi chuyển qua bài tìm x là xong, hơi dài đấy ^^

Học tốt ^^

\(\frac{1}{1.5}+\frac{1}{5.9}+\frac{1}{9.13}+\frac{1}{13.17}+...+\frac{1}{41.45}\)

\(=\frac{1}{4}.\left(1-\frac{1}{5}+\frac{1}{5}-\frac{1}{9}+\frac{1}{9}-\frac{1}{13}+...+\frac{1}{41}-\frac{1}{45}\right)\)

\(=\frac{1}{4}.\left(1-\frac{1}{45}\right)\)

\(=\frac{1}{4}.\frac{44}{45}\)

\(=\frac{11}{45}\)

Đặt \(\frac{1}{1\cdot5}+\frac{1}{5\cdot9}+\frac{1}{9\cdot13}+\frac{1}{13\cdot17}+...+\frac{1}{41\cdot45}\)  là A.

Ta có:

\(A=\frac{1}{1\cdot5}+\frac{1}{5\cdot9}+\frac{1}{9\cdot13}+\frac{1}{13\cdot17}+...+\frac{1}{41\cdot45}\)

\(4A=4\left(\frac{1}{1\cdot5}+\frac{1}{5\cdot9}+\frac{1}{9\cdot13}+\frac{1}{13\cdot17}+...+\frac{1}{41\cdot45}\right)\)

\(4A=\frac{4}{1\cdot5}+\frac{4}{5\cdot9}+\frac{4}{9\cdot13}+\frac{4}{13\cdot17}+...+\frac{4}{41\cdot45}\)

\(4A=1-\frac{1}{5}+\frac{1}{5}-\frac{1}{9}+\frac{1}{9}-\frac{1}{13}+\frac{1}{13}-\frac{1}{17}+...+\frac{1}{41}-\frac{1}{45}\)

\(4A=1-\frac{1}{45}\)

\(4A=\frac{44}{45}\)

\(A=\frac{44}{45}:4\)

\(A=\frac{11}{45}\)

Vậy \(\frac{1}{1\cdot5}+\frac{1}{5\cdot9}+\frac{1}{9\cdot13}+\frac{1}{13\cdot17}+...+\frac{1}{41\cdot45}=\frac{11}{45}\)

cảm ơn 2 bn nhé

mk kết bạn nha

\(\frac{1}{1.5}+\frac{1}{5.9}+\frac{1}{9.13}+...+\frac{1}{97.101}\)

\(=\frac{1}{4}\left(1-\frac{1}{5}+\frac{1}{5}-\frac{1}{9}+\frac{1}{9}-\frac{1}{13}+...+\frac{1}{97}-\frac{1}{101}\right)\)

\(=\frac{1}{4}.\left(1-\frac{1}{101}\right)\)

\(=\frac{1}{4}.\frac{100}{101}\)

\(=\frac{25}{101}\)

1/1.5+/5.9+1/9.13..........+1/101.103

=1-1/5+1/5-1/7+1/9-1/13.........+1/101-1/103

=1-1/103

=102/103

XIN 5 TÍCH VÌ MẤT 5 PHÚT

OK

\(A=\frac{1}{1.5}+\frac{1}{5.9}+\frac{1}{9.13}+...+\frac{1}{93.97}\)

\(A=\frac{1}{4}.\left(\frac{4}{1.5}+\frac{4}{5.9}+\frac{4}{9.13}+...+\frac{1}{93.97}\right)\)

\(A=\frac{1}{4}.\left(1-\frac{1}{5}+\frac{1}{5}-\frac{1}{9}+\frac{1}{9}-\frac{1}{13}+...+\frac{1}{93}-\frac{1}{97}\right)\)

\(A=\frac{1}{4}.\left(1-\frac{1}{97}\right)\)

\(A=\frac{1}{4}.\frac{96}{97}=\frac{24}{97}\)

\(A=\frac{1}{1.5}+\frac{1}{5.9}+\frac{1}{9.13}+...+\frac{1}{93.97}\)

\(A=\frac{1}{4}.\left(\frac{4}{1.5}+\frac{4}{5.9}+\frac{4}{9.13}+...+\frac{1}{93.97}\right)\)

\(A=\frac{1}{4}.\left(1-\frac{1}{5}+\frac{1}{5}-\frac{1}{9}+\frac{1}{9}-\frac{1}{13}+...+\frac{1}{93}-\frac{1}{97}\right)\)

\(A=\frac{1}{4}.\left(1-\frac{1}{97}\right)\)

\(A=\frac{1}{4}.\frac{96}{97}=\frac{24}{97}\)