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

\(A=\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{1}{16}+\frac{1}{32}\)

=>  \(2A=1+\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{1}{16}\)

=>  \(2A-A=\left(1+\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{1}{16}\right)-\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{1}{16}+\frac{1}{32}\right)\)

=>  \(A=1-\frac{1}{32}=\frac{31}{32}\)

ok minh se giup vao 123 nha

bai1 tu -1738 den -16 co so cac so nguyen am la : (1738-16)/2+1=862

C1

B=1+3+5+..+99=50.(99+1)/2=50.50=2500

C=-(2+4+6+...+100)=50.(100+2)/2=50*51=-2550

A=B+C=2500-2550=-50

C2

A=1+(3-2)+(5-4)+...+(99-98)+100

=1+1+1+...+1-100

=1+49-100=-50

\(A=1-2+3-4+...+97-98+99\)

\(A=\left(1-2\right)+\left(3-4\right)+...+\left(97-98\right)+99\)

\(A=-1+\left(-1\right)+...+\left(-1\right)+99\)

\(A=\left(-1\right).49+99\)

\(A=-49+99\)

\(A=50\)

A = 1 - 2 + 3 - 4 +…+ 97 - 98 + 99

A = ( 1 - 2 ) + ( 3 - 4 ) + ... + ( 97 - 98 ) + 99

A = ( -1 ) + ( -1 ) + ... + ( -1 ) + 99

A = ( -1 ) . 49 + 99

A = 50