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a) A =1+3+32+33+...+3100
3A = 3 + 32+33+...+3101
3A-A=( 3 + 32+33+...+3101)-(1+3+32+33+...+3100)
2A = 3101-1
A = \(\frac{3^{101}-1}{2}\)
Thùy An làm sai rùi
Lời giải:
a) \(A=1+3+3^2+3^3+...+3^{100}\)
\(\Rightarrow 3A=3+3^2+3^3+...+3^{101}\)
Trừ theo vế:
\(\Rightarrow 3A-A=(3+3^2+3^3+..+3^{101})-(1+3+3^2+...+3^{100})\)
\(2A=3^{101}-1\Rightarrow A=\frac{3^{101}-1}{2}\)
b) \(B=2^{100}-2^{99}+2^{98}-2^{97}+...+2^2-2\)
\(\Rightarrow 2B=2^{101}-2^{100}+2^{99}-2^{98}+...+2^3-2^2\)
Cộng theo vế:
\(\Rightarrow B+2B=2^{201}-2\)
\(\Rightarrow B=\frac{2^{101}-2}{3}\)
c) Ta có:
\(C=3^{100}-3^{99}+3^{98}-3^{97}+...+3^2-3+1\)
\(\Rightarrow 3C=3^{101}-3^{100}+3^{99}-3^{98}+...+3^3-3^2+3\)
Cộng theo vế:
\(C+3C=(3^{100}-3^{99}+3^{98}-....+3^2-3+1)+(3^{101}-3^{100}+3^{99}-....+3^3-3^2+3)\)
\(4C=3^{101}+1\Rightarrow C=\frac{3^{101}+1}{4}\)
a: \(3A=3+3^2+...+3^{101}\)
\(\Leftrightarrow2A=3^{101}-1\)
hay \(A=\dfrac{3^{101}-1}{2}\)
b: \(2B=2^{101}-2^{100}+...+2^3-2^2\)
\(\Leftrightarrow3B=2^{101}-2\)
hay \(B=\dfrac{2^{101}-2}{3}\)
c: \(3C=3^{101}-3^{100}+....+3^3-3^2+3\)
=>\(4C=3^{101}+1\)
hay \(C=\dfrac{3^{101}+1}{4}\)
b) B = 2100 - 299 + 298 - 297 + ...+ 22 - 2
=> B x 2 = 2101 - 2100 + 299 - 298 + ...23 - 22
=> B x 2 + B = (2101 - 2100 + 299 - 298 + ...23 - 22 ) + (2100 - 299 + 298 - 297 + ...+ 22 - 2)
<=> B x 3 = 2101 - 2 = 2. ( 299 - 1)
=> B = \(\frac{2.\left(2^{99}-1\right)}{3}\)
Phần c) Làm tương tự Lấy C x 3 rồi + với C.
\(A=2^{100}-2^{99}+2^{98}-2^{97}+....+2^2-2\)
\(2A=2^{101}-2^{100}+2^{99}-2^{98}+....+2^3-2^2\)
\(2A+A=2^{101}-2\)
\(A=\frac{2^{101}-2}{3}\)
b) tương tự
\(B=\frac{3^{101}+1}{4}\)
A = 2100 - 299 + 298 - 297 + ... + 22 - 2
= ( 2100 + 298 + ... + 22 ) - ( 299 + 297 + ... + 2 )
= ( 2100 + 298 + ... + 22 ) - 2( 299 + 297 + ... + 2 ) + ( 299 + 297 + ... + 2 )
= 299 + 297 + ... + 2
=> 4A = 2103 + 299 + ... + 23
=> 3A = 2103 - 2
=> A = \(\frac{2^{103}-2}{3}\)
a) \(A=2^{100}-2^{99}+2^{98}-2^{97}+...+2^2-2\)
\(2A=2^{101}-2^{100}+2^{99}-2^{98}+...+2^3-2^2\)
\(\Rightarrow A+2A=2^{101}-2\)
\(A\left(1+2\right)=2^{101}-2\)
\(A.3=2^{101}-2\)
\(A=\frac{2^{101}-2}{3}\)
b) \(B=3^{100}-3^{99}+3^{98}-3^{97}+...+3^2-3\)
\(3B=3^{101}-3^{100}+3^{99}-3^{98}+...+3^3-3^2\)
\(\Rightarrow B+3B=3^{101}-3\)
\(B\left(1+3\right)=3^{101}-3\)
\(4B=3^{101}-3\)
\(B=\frac{3^{101}-3}{4}\)
A = 2100 - 299 + 298 - 297 +...+ 22 - 2
=> 2A = 2101 - 2100+299 - 298+...+23-22
=> 2A+A= 2101 -2
=> \(A=\frac{2^{101}-2}{3}\)
phần B bn lm tương tự nha!