a, 1 + 2\(^2\) + 2\(^3\)+ ... + 2\(^{100}\) b, 2 + 2\(^2\)+ 2\(^3\)+ ... + 2\(^{100}\)
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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) 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
2:
\(B=\left(\dfrac{1}{2^2}-1\right)\left(\dfrac{1}{3^2}-1\right)\cdot...\cdot\left(\dfrac{1}{100^2}-1\right)\)
\(=\left(\dfrac{1}{2}-1\right)\left(\dfrac{1}{2}+1\right)\left(\dfrac{1}{3}-1\right)\left(\dfrac{1}{3}+1\right)\cdot...\cdot\left(\dfrac{1}{100}-1\right)\left(\dfrac{1}{100}+1\right)\)
\(=\left(\dfrac{1}{2}-1\right)\left(\dfrac{1}{3}-1\right)\cdot...\cdot\left(\dfrac{1}{100}-1\right)\left(\dfrac{1}{2}+1\right)\left(\dfrac{1}{3}+1\right)\cdot...\cdot\left(\dfrac{1}{100}+1\right)\)
\(=\dfrac{-1}{2}\cdot\dfrac{-2}{3}\cdot...\cdot\dfrac{-99}{100}\cdot\dfrac{3}{2}\cdot\dfrac{4}{3}\cdot...\cdot\dfrac{101}{100}\)
\(=-\dfrac{1}{100}\cdot\dfrac{101}{2}=\dfrac{-101}{200}< -\dfrac{100}{200}=-\dfrac{1}{2}\)
Bài 1:
A = 1 + 3 + 32 + ... + 3100
=> 3A = 3 + 32 + ... + 3101
=> 2A = 3101 - 1
=> A = \(\frac{3^{101}-1}{2}\)
B = 1 + 42 + 44 + ... + 4100
=> 8B = 42 + 44 + ... + 4102
=> 7B = 4102 - 1
=> B = \(\frac{4^{102}-1}{7}\)
Bài 2:
a) S1 = 22 + 42 + ... + 202
=> S1 = 22(1+22+...+102)
=> S1 = 22.385
=> S1 = 1540
b) S2 = 1002 + 2002 + ... + 10002
=> S2 = 1002(1+22+...+102)
=> S2 = 1002.385
=> S2 = 3850000
A = 12 + 22 + 32 + ........ + 1002
= 1(2 - 1) + 2(3 - 1) + 3(4 - 1) + ........ + 100(101 - 1)
= 1.2 - 1 + 2.3 - 2 + 3.4 - 3 + ......... + 100.101 - 100
= (1.2 + 2.3 + 3.4 + ........ + 99.100) + 100.101 - (1 + 2 + 3 + ....... + 100)
= B + 10100 - \(\frac{100.101}{2}\) = B + 5050
=> A - B = 5050
a) đặt \(M=1+2^2+2^3+...+2^{100}\)
\(\Rightarrow2M=2+2^3+2^4+...+2^{101}\)
\(\Rightarrow2M-M=\left(2+2^3+2^4+...+2^{101}\right)-\left(1+2^2+2^3+...+2^{100}\right)\)
\(\Rightarrow M=2-1+2^{101}=1+2^{101}\)
b) đặt \(N=2+2^2+2^3+...+2^{100}\)
\(\Rightarrow2N=4+2^3+2^4+...+2^{101}\)
\(\Rightarrow2N-N=\left(4+2^3+2^4+...+2^{101}\right)-\left(2+2^2+2^3+...+2^{100}\right)\)
\(\Rightarrow N=4-2-2^2+2^{101}=-2+2^{101}\)