A=2100-299+298-297+.....+22-2
=>2A=2101-2100+299-298+.....+23-22
=>2A+A=2101-2100+299-298+.....+23-22+2100-299+298-297+....+22-2
=>3A=2201-2
=>A=\(\frac{2^{201}-2}{3}\)
B=3100-399+398-397+....+32-3+1
=>3B=3101-3100+399-398+...+33-32+3
=>3B+B=3101-3100+399-398+...+33-32+3+3100-399+398-397+....+32-3+1
=>4B=3101+1
=>B=\(\frac{3^{101}+1}{4}\)
Câu a : Đặt 2A = 2^101 - 2^100 + 2^99 - 2^98 +...+ 2^3 - 2^2
=> 2A - A = 2^101 - 2^100 + 2^99 - 2^98 +...+ 2^3 - 2^2 - ( 2^100 - 2^99 + 2^98 - 2^97 +...+ 2^2 - 2)
=> A = 2^101 - 2^100 + 2^99 - 2^98 +...+ 2^3 - 2^2 - 2^100 + 2^99 - 2^98 + 2^97 -...- 2^2 + 2
=> A= = 2^101 -2(2^100 + 2^98 + 2^96 +...+ 2^2) + 2(2^99 + 2^97 + 2^95 +...+ 2^3) +2
Câu b : Làm tương tự như trên
BẤM ĐÚNG CHO MÌNH NHA
Câu a : Đặt 2A = 2^101 - 2^100 + 2^99 - 2^98 +...+ 2^3 - 2^2
=> 2A - A = 2^101 - 2^100 + 2^99 - 2^98 +...+ 2^3 - 2^2 - ( 2^100 - 2^99 + 2^98 - 2^97 +...+ 2^2 - 2)
=> A = 2^101 - 2^100 + 2^99 - 2^98 +...+ 2^3 - 2^2 - 2^100 + 2^99 - 2^98 + 2^97 -...- 2^2 + 2
=> A= = 2^101 -2(2^100 + 2^98 + 2^96 +...+ 2^2) + 2(2^99 + 2^97 + 2^95 +...+ 2^3) +2
Câu b : Làm tương tự như trên
Toán lớp 6 đấy các bồ bấm đúng cho tui nhé!!!
bạn trieu dang viết nhầm rồi.Đáng lẽ A=2^101-2/3 chứ
M = 2^{100-}2^{99}+2^{98}-...+2^2-22100−299+298−...+22−2
2M=2^{101}-2^{100}+2^{99}-2^{98}+2^{97}+...+2^3-2^22M=2101−2100+299−298+297+...+23−22
2M+M=2^{101}-22M+M=2101−2
M=\frac{2^{101}-2}{3}M=32101−2
B=3100−399+...+32−3+1
3B=3\left(3^{100}-3^{99}+...+3^2-3+1\right)3B=3(3100−399+...+32−3+1)
3B=3^{101}-3^{100}+...+3^3-3^2+33B=3101−3100+...+33−32+3
3B+B=\left(3^{101}-3^{100}+...+3^3-3^2+3\right)+\left(3^{100}-3^{99}+...+3^2-3+1\right)3B+B=(3101−3100+...+33−32+3)+(3100−399+...+32−3+1)
4B=3^{101}+1\Rightarrow B=\frac{3^{101}+1}{4}4B=3101+1⇒B=43101+1
\(2A=2^{101}-2^{100}+2^{99}-2^{98}+...+2^3-2^2\)
\(2A+A=2^{101}-2\)
\(A=\frac{2^{101}-2}{3}\)
