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A=1+2+22+…+2100
2A=2(1+2+22+…+2100)
2A=2+22+…+2101
2A-A = A = 2+22+…+2101-(1+2+22+…+2100)
A = 2+22+…+2101-1-2-22-…-2100
A = (2-2)+(22-22)+…+(2100-2100)+2101-1
A = 0+0+…+0+2101-1
A = 2101-1
B=3-32+33-34+…+299-3100
3B = 3(3-32+33-34+…+299-3100)
3B = 32-33+34-…-299+3100-3101
3B+B = 4B = 3-32+33-34+…+299-3100
4B =(3-32+33-34+…+299-3100)+(32-33+34-…-299+3100-3101)
4B =3-32+33-34+…+299-3100+32-33+34-…-299+3100-3101
4B =3+(32-32)+(33-33)+(34-34)+…+(299-299)+(3100-3100)-3101
4B =3+0+0+0+....+0-3101
4B =3-3101
B = (3-3101)/4
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A = 1 . 2 + 2 . 3 + 3 . 4 + ... + 99 .100
3 . A = 1. 2 . 3 + 2 . 3 . 3 + 3 . 4 . 3 + ... + 99 . 100 . 3
3 . A = 1 . 2 . 3 + 2 . 3 . ( 4 - 1 ) + 3 . 4 . ( 5 - 2 ) + ... + 99 . 100 . ( 1001 - 998 )
3 . A = 1 . 2 . 3 + 2 . 3 . 4 - 1 . 2 . 3 + 3 . 4 . 5 - 2 . 3 . 4 + ... + 99 . 100 . 1001 - 998 . 99 . 100
3 . A = 99 . ( 100 . 10 )
A = ( 99 . 100 . 10 ) : 3
A = 33000
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Bài 1:
a: \(2A=2^{101}+2^{100}+...+2^2+2\)
\(\Leftrightarrow A=2^{100}-1\)
b: \(3B=3^{101}+3^{100}+...+3^2+3\)
\(\Leftrightarrow2B=3^{100}-1\)
hay \(B=\dfrac{3^{100}-1}{2}\)
c: \(4C=4^{101}+4^{100}+...+4^2+4\)
\(\Leftrightarrow3C=4^{101}-1\)
hay \(C=\dfrac{4^{101}-1}{3}\)
\(A=3+3^2+...+3^{100}\)
\(\Rightarrow3A=3^2+3^3+...+3^{101}\)
\(\Rightarrow3A-A=\left(3^2+3^3+...+3^{101}\right)-\left(3+3^2+...+3^{100}\right)\)
\(\Rightarrow2A=3^{101}-3\)
\(\Rightarrow A=\frac{3^{101}-3}{2}\)
Ta có: A = 31 + 32 + 33 + … + 399+3100