a: \(\dfrac{1}{2}+\dfrac{-3}{8}+\dfrac{5}{9}\)

\(=\dfrac{36}{72}-\dfrac{27}{72}+\dfrac{40}{72}\)

\(=\dfrac{49}{72}\)

b: \(\dfrac{13}{-30}+\dfrac{17}{45}+\dfrac{-7}{18}\)

\(=\dfrac{-13}{30}+\dfrac{17}{45}+\dfrac{-7}{18}\)

\(=\dfrac{-39+34-35}{90}=\dfrac{-40}{90}=-\dfrac{4}{9}\)

c: \(-\dfrac{5}{19}+\dfrac{7}{4}-\left(-\dfrac{2}{3}\right)+\dfrac{5}{6}\)

\(=-\dfrac{5}{19}+\dfrac{7}{4}+\dfrac{2}{3}+\dfrac{5}{6}\)

\(=\dfrac{-5}{19}+\dfrac{21+8+10}{12}\)

\(=-\dfrac{5}{19}+\dfrac{39}{12}=\dfrac{-5}{19}+\dfrac{13}{4}=\dfrac{227}{76}\)

Số số hạng trong tổng trên là:

( n -1 ) : 1 + 1 = n -1

Tổng trên có giá trị là

( n + 1 ) . ( n - 1 ) : 2 = ( n² - 1 ) : 2

= ( n + 1 ) . n : 2

Thông cảm nha >

Mình giải nhầm

Đáp án đúng ở câu hỏi của cậu sau ấy

um cam on ban nha

Ta có : a) A= 1+ 5+ 5²+ 5³+........+ 5¹⁹⁹⁸ 

=> 5A =  5+ 5²+ 5³+........+ 5¹⁹⁹⁹

=> 5A - A = 5¹⁹⁹⁹ - 1

=> 4A  = 5¹⁹⁹⁹ - 1

\(\Rightarrow A=\frac{5^{1999}-1}{4}\)

b) Ta có : b) B= 1+ 4+ 4²  + ...... + 4ⁿ 

=> 4B = 4 + 4²  + 4³ + ...... + 4^{n + 1} 

=> 4B - B = 4^{n + 1} - 1

=> 3B = 4^{n + 1} - 1

=> \(B=\frac{4^{n+1}-1}{3}\)

Sì tin

đề câu e sai r

a) 1+2+3+4+5+...+n = n(n+1) / 2

b)2+4+6+...+2n = [(2n-2):2+1] . (2n+2)/2 = n . ( 2n+2) /2

 mấy câu còn lại bn tự tính đi

mik lười quá :))

a) \(1+2+3+4+...+n\)

\(=\left(n+1\right)\left[\left(n-1\right):1+1\right]:2\)

\(=\left(n+1\right)\left(n-1+1\right):2\)

\(=n\left(n+1\right):2\)

\(=\dfrac{n\left(n+1\right)}{2}\)

b) \(2+4+6+..+2n\)

\(=\left(2n+2\right)\left[\left(2n-2\right):2+1\right]:2\)

\(=2\left(n+1\right)\left[2\left(n-1\right):2+1\right]:2\)

\(=\left(n+1\right)\left(n-1+1\right)\)

\(=n\left(n+1\right)\)

c) \(1+3+5+...+\left(2n+1\right)\)

\(=\left[\left(2n+1\right)+1\right]\left\{\left[\left(2n-1\right)-1\right]:2+1\right\}:2\)

\(=\left(2n+1+1\right)\left[\left(2n-1-1\right):2+1\right]:2\)

\(=\left(2n+2\right)\left[\left(2n-2\right):2+1\right]:2\)

\(=2\left(n+1\right)\left[2\left(n-1\right):2+1\right]:2\)

\(=\left(n+1\right)\left(n-1+1\right)\)

\(=n\left(n+1\right)\)

d) \(1+4+7+10+...+2005\)

\(=\left(2005+1\right)\left[\left(2005-1\right):3+1\right]:2\)

\(=2006\cdot\left(2004:3+1\right):2\)

\(=2006\cdot\left(668+1\right):2\)

\(=1003\cdot669\)

\(=671007\)

e) \(2+5+8+...+2006\)

\(=\left(2006+2\right)\left[\left(2006-2\right):3+1\right]:2\)

\(=2008\cdot\left(2004:3+1\right):2\)

\(=1004\cdot\left(668+1\right)\)

\(=1004\cdot669\)

\(=671676\)

g) \(1+5+9+...+2001\)

\(=\left(2001+1\right)\left[\left(2001-1\right):4+1\right]:2\)

\(=2002\cdot\left(2000:4+1\right):2\)

\(=1001\cdot\left(500+1\right)\)

\(=1001\cdot501\)

\(=501501\)