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a, Đặt \(A=1+3+3^2+3^3+....+3^{100}\)
=> \(3A=3+3^2+3^3+3^4+...+3^{101}\)
=> \(2A=3A-A=3^{101}-1\)
=> \(A=\frac{3^{101}-1}{2}\)
Vậy giá trị của biểu thức là \(\frac{3^{101}-1}{2}\)
b, Đặt \(B=1+4+4^2+2^3+....+4^{50}\)
=> \(4B=4+4^2+4^3+4^4+....+4^{51}\)
=> \(3B=4B-B=4^{51}-1\)
=> \(B=\frac{4^{51}-1}{3}\)
Vậy giá trị của biểu thức là \(\frac{4^{51}-1}{3}\)
Bài 1
a) A = 2^0 + 2^1 + 2^2 +...+ 2^50
2A=2^1+2^2+2^3+...+2^51
2A-A=(2^1+2^2+2^3+...+2^51)-(2^0 + 2^1 + 2^2 +...+ 2^50)
A=(2^1-2^1)+(2^2-2^2)+...+(2^50-2^50)+(2^51-2^1)
A=0+0+...+0+(2^51-2^1)
A=2^51-2^1
b)B = 5 + 5^2 + 5^3 +...+ 5^99 + 5^100
5B=5^2+5^3+5^4+...+5^100+5^101
5B-B=(5^2+5^3+5^4+...+5^100+5^101)-( 5 + 5^2 + 5^3 +...+ 5^99 + 5^100)
4B=(5^2-5^2)+(5^3-5^3)+...+(5^100-5^100)+(5^101-5)
4B=0+0+...+0+(5^101-5)
4B=5^101-5
B=(5^101-5)/4
c)C = 3 - 3^2 + 3^3 - 3^4 +...+ 3^2009 - 3 ^2010
3C=3^2-3^3+3^4-3^5+...+3^2010-3^2011
3C-C=(3^2-3^3+3^4-3^5+...+3^2010-3^2011)-(3 - 3^2 + 3^3 - 3^4 +...+ 3^2009 - 3 ^2010)
...............................................!!!!!!!!!!!!!!!!!!!!!!!!
Bài 2
8(mình k0 chắc)
A=1+4+42+43+44+........+42015
4A=4(A=1+4+42+43+44+........+42015)
4A=4+42+43+44+45+........+42016
4A-A=(4+42+43+44+45+........+42016)-(1+4+42+43+44+........+42015)
3A=42016-1
A=(42016-1):3
a, (-45)+(-2018)+/-145/+/-2018/
=[(-45)+/-45/] +[(-2018)+/-2018/)
=(-45+45)+(-2018+2018)
=0
b, 5^3.4^3+85.5^3-5^7:5^4
=5^3.4^3+85.5^3-5^3
=5^3.(4^3+85-1)
=5^3.148
=18500
c, (7^108.50-7^108):7^110
=[7^108.(50-1)];7^110
=(49.7^108):7^110
=49:7^2
=49:49=1
a) (-45)+(-2018)+/-145/+/-2018/
=(-45)+(-2018)+145+2018
=(-45+145)+(-2018+2018)
=100+0
=100
b)53.42+85.53-57:54
=53.42+85.53-53
=53.(42+85-1)
=53.100
=12500
c) (7108.50-7108):7110
=7108.(50-1):7110
=7108.49:7110
=7108.72:7110
=7110:7110
=1
\(B=3+3^2+3^3+3^4+...+3^{50}\)
\(\Rightarrow3B=3^2+3^3+3^4+3^5+...+3^{51}\)
\(\Rightarrow2B=3^{51}-3\)
\(\Rightarrow B=\frac{3^{51}-3}{2}\)
\(C=4+4^2+4^3+4^4+...+4^{2018}\)
\(\Rightarrow4C=4^2+4^3+4^4+4^5+...+4^{2019}\)
\(\Rightarrow3C=4^{2019}-4\)
\(\Rightarrow C=\frac{4^{2019}-4}{3}\)
\(B=3+3^2+3^3+...+3^{50}\)
\(\Rightarrow3B=3^2+3^3+3^4+....+3^{51}\)
\(\Rightarrow3B-B=\left(3^2+3^3+3^4+...+3^{51}\right)-\left(3+3^2+...+3^{50}\right)\)
\(\Rightarrow2B=3^{51}-3\)
\(\Rightarrow B=\frac{3^{51}-3}{2}\)
\(C=4+4^2+4^3+...+4^{2018}\)
\(\Rightarrow4C=4^2+4^3+4^4+....+4^{2019}\)
\(\Rightarrow4C-C=\left(4^2+4^3+4^4+...+4^{2019}\right)-\left(4+4^2+4^3+...+4^{2018}\right)\)
\(\Rightarrow3C=4^{2019}-4\)
\(\Rightarrow C=\frac{4^{2019}-4}{3}\)