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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)
a, A = 1 + 3 + 3\(^{^2}\) + .... + 3\(^{100}\)
3A = 3 + 3\(^2\) + ..... + 3\(^{101}\)
Lấy 3A - A
\(\Rightarrow\) 2A = 3\(^{101}\) - 1
A = \(\frac{3^{101}-1}{2}\)
b, Áp dụng kiến thức câu a
1/2.A=1/22+1/23+...+1/2101
=>1/2A-A=1/2101-1/2
=>-1/2A=1/2101-1/2
A=(1/2101-1/2):(-1/2)=(1/2101-1/2).(-2)
=1-1/2100
\(C=\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+...+\frac{1}{3^{98}}+\frac{1}{3^{99}}\)
\(3C=1+\frac{1}{3}+\frac{1}{3^2}+...+\frac{1}{3^{97}}+\frac{1}{3^{98}}\)
\(3C-C=\left(1+\frac{1}{3}+\frac{1}{3^2}+...+\frac{1}{3^{97}}+\frac{1}{3^{98}}\right)-\left(\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+...+\frac{1}{3^{98}}+\frac{1}{3^{99}}\right)\)
\(2C=1-\frac{1}{3^{99}}< 1\)
\(\Rightarrow C=\frac{1-\frac{1}{3^{99}}}{2}< \frac{1}{2}\)
1.
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 = \(\frac{3^{101}+1}{4}\)
a)2A=4+4^2+4^3+...+4^101
2A-A=4^101-1
A=4^101-1
khong bit phai hoi muon gioi phai hoc
\(2A=2+1+\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^{2016}}\)
\(2A-A=\left(2+1+\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^{2016}}\right)-\left(1+\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+\frac{1}{2^4}+...+\frac{1}{2^{2017}}\right)\)
\(A=2-\frac{1}{2^{2017}}\)
A=1+1/2+1/22+1/23+...+1/22017
1/2A=1/2+1/22+1/23+1/24+...+1/22018
A-1/2A=(1+1/2+1/22+1/23+...+1/22017)-(1/2+1/22+1/23+1/24+...+1/22018)
A-1/2A=1-22012018
1/2A=1-1/22018
A=(1-1/22018).2
A=2-22019
\(C=1+\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^{100}}\)
\(2C=2+1+\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^{99}}\)
\(2C-C=\left(2+1+\frac{1}{2}+\frac{1}{2^2}+...+2^{99}\right)-\left(1+\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^{100}}\right)\)
\(\Rightarrow C=2-\frac{1}{2^{100}}\)
Ủng hộ mk nha bn !!! ^_^
2C= 2+1+1/2+1/22+.....+1/299
2C-C= ( 2+1+1/2+1/22+.....+1/299 ) - (1+1/2+1/22+1/23+.....+1/2100)
C= 2 - 1/2100
C=2