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a)
A = 2 + 22 + 23 + 24 + .... +299 + 2100
2A = 22 + 23 + 24 + 25 + ... + 2100 + 2101
2A - A = A = 2101 - 2
vậy A = 2101 - 2
b)
B = 1 + 2 + 22 + 23 + ... + 22017
2B = 2 + 22 + 23 + 24 + ... + 22018
2B - B = B = 22018 - 1
Vậy B = 22018 - 1
c)
C = 2 + 23 + 25 + ... + 22017
4C = 23 + 25 + 27 + ... + 22019
4C - C = 3C = 22019 - 2
C = \(\frac{2^{2019}-2}{3}\)
d)
D = 2 + 24 + 27 + ... + 22017
8D = 24 + 27 + 210 + ... + 22020
8D - D = 7D = 22020 - 2
D = \(\frac{2^{2020}-2}{7}\)
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c.
C= ( a+b+c)2+(a+b-c)2- 2(a+b)2
=a2+b2+c2+a2+b2- c2-2a2-2b2
= 2a2+2b2+c2-c2-2a2-2b2
= 0
Vậy C= 0
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phần b tương tự phần a nên em làm câu a và c thôi :
a, \(M=1-2+2^2-2^3+...+2^{2012}\)
\(2M=2-2^2+2^3-2^4+...+2^{2013}\)
\(3M=2^{2013}+1\)
\(M=\frac{2^{2013}+1}{3}\)
c, \(E=2^{100}-2^{99}-2^{98}-...-1\)
\(E=2^{100}-\left(2^{99}+2^{98}+...+1\right)\)
đặt \(A=2^{99}+2^{98}+...+1\)
\(2A=2^{100}+2^{98}+...+2\)
\(2A-A=2^{100}-1\) hay \(A=2^{100}-1\)
ta có :
\(E=2^{100}-\left(2^{100}-1\right)\)
\(E=2^{100}-2^{100}+1=1\)
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A= 1/2+1/22+1/23+1/24+.....+1/22019
2A= 1+1/2+1/22+1/23+1/24+.....+1/22018
2A-A=(1+1/2+1/22+1/23+1/24+.....+1/22018)-(1/2+1/22+1/23+1/24+.....+1/22019)
A=1-1/22019
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A=1+2+22+23+...+263
2A=2+22+23+...+263+264
\(-\)
\(A=1+2+2^2+....+2^{63}\)
\(A=2^{64}-1\)
Vậy A=264-1
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a/ta gọi biểu thức trên là A.
ta có: A=1+2+22+...+2100
2A= 2x(1+2+22+...+2100)
2A= 2x1+2x2+22x2+...+2100x2
2A= 2+22+23+....+2101
2A-A=A=(2+22+23+....+2101)-(1+2+22+...+2100)
A= 2101-1
b/ làm tương tụ như câu a nhưng cuối cùng phải thêm '':2'' (vì lúc đó ta tính ra 3A - A =2A nên phải chia 2)
\(B=1+2^1+2^2+2^3+2^4+...+2^{100}\\ 2\text{B}=2^1+2^2+2^3+2^4+2^5+...+2^{101}\\ 2\text{B}-B=b=\left(2^1+2^2+2^3+2^4+2^5+...+2^{101}\right)-\left(1+2^1+2^2+2^3+2^4+...+2^{100}\right)\\ B=2^{101}-1\)