a,
A = 4 + 22 + 23 + 24 + .. + 220
Đặt A1 = 22 + 23 + 24 + .. + 220
2A1 = 2.( 22 + 23 + 24 + .. + 220)
= 23 + 24 + 25 + ... + 22
2A1 - A1 = (22 + 23 + 24 + .. + 220) - (23 + 24 + 25 + ... + 22 )
A1 = 221 - 22
= 221 - 4
=> A = 4 + 221 - 4
=> A = 221
b,
(x + 1) + (x + 2) + (x + 3) + .... + (x + 100) = 5750
=> 100x + (1 + 2 + 3 + ... + 100) = 5750
=> 100x + 5050 = 5750
=> 100x = 5750 - 5050
=> 100x = 700
=> x = 700 : 100
=> x = 7
a)Ta có : \(A=4+2^2+2^3+2^4+...+2^{20}\)
\(\Rightarrow2A=2^3+2^3+2^4+2^5+...+2^{21}\)
\(\Rightarrow2A-A=\left(2^3+2^3+2^4+2^5+...+2^{21}\right)-\left(4+2^2+2^3+2^4+...+2^{20}\right)\)
hay \(A=2^{21}+2^3-2^2-4=2^{21}+2^3-8=2^{21}\)
\(\Rightarrowđpcm\)
b) Ta có: \(\left(x+1\right)+\left(x+2\right)+\left(x+3\right)+...+\left(x+100\right)=5750\)
\(\Rightarrow x+1+x+2+x+3+...+x+100=5750\)
\(\Rightarrow100x+\left(1+2+3+...+100\right)=5750\)
\(\Rightarrow100x+\frac{100.101}{2}=5750\)
\(\Rightarrow100x+5050=5750\)
\(\Rightarrow100x=700\)
\(\Rightarrow x=7\)
Vậy \(x=7\)
c) Ta có : \(A=x+8=\left(x+3\right)+5\)
\(\Rightarrow\)Để \(A⋮x+3\)
thì \(\left(x+3\right)+5⋮x+3\)
mà \(x+3⋮x+3\)
\(\Rightarrow5⋮x+3\)
Vì \(x\inℕ\Rightarrow x+3\inℕ\Rightarrow x+3\inƯ\left(5\right)=\){\(\pm1;\pm5\)}
Nhưng \(x\ge0\Rightarrow x+3\ge3\Rightarrow x+3=5\)
\(\Rightarrow x=2\)
Vậy \(x=2\)
Câu a tui ko bt nhưng câu b đây:
(x + 1) + (x+2) + (x+3) +...+(x+100) = 5750
=x + 1+ x+2+ x+3+...+ x+100
= \(\left(100x\right)+\)SSH từ 1 đến 100 = 5750
=\(100x+\left(1+100\right).100.100:2=5750\)
= \(100x+5050=5750\)
=\(100x=5750-5050\)
\(100=700\)
\(x=700:100\)
\(x=7\)
c)
Theo tui thì \(x=\frac{A}{9}-\frac{1}{3}\)
Phần c) bỏ qua nha :)