B = 1+3+3^2+3^3+......+3^100
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NN
0
14 tháng 11 2023
2:
\(B=\left(\dfrac{1}{2^2}-1\right)\left(\dfrac{1}{3^2}-1\right)\cdot...\cdot\left(\dfrac{1}{100^2}-1\right)\)
\(=\left(\dfrac{1}{2}-1\right)\left(\dfrac{1}{2}+1\right)\left(\dfrac{1}{3}-1\right)\left(\dfrac{1}{3}+1\right)\cdot...\cdot\left(\dfrac{1}{100}-1\right)\left(\dfrac{1}{100}+1\right)\)
\(=\left(\dfrac{1}{2}-1\right)\left(\dfrac{1}{3}-1\right)\cdot...\cdot\left(\dfrac{1}{100}-1\right)\left(\dfrac{1}{2}+1\right)\left(\dfrac{1}{3}+1\right)\cdot...\cdot\left(\dfrac{1}{100}+1\right)\)
\(=\dfrac{-1}{2}\cdot\dfrac{-2}{3}\cdot...\cdot\dfrac{-99}{100}\cdot\dfrac{3}{2}\cdot\dfrac{4}{3}\cdot...\cdot\dfrac{101}{100}\)
\(=-\dfrac{1}{100}\cdot\dfrac{101}{2}=\dfrac{-101}{200}< -\dfrac{100}{200}=-\dfrac{1}{2}\)
VT
1 tháng 5 2017
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
25 tháng 6 2021
a) Ta có: \(A=1^3+2^3+3^3+...+100^3\)
\(=\left(1-1\right)\cdot1\cdot\left(1+1\right)+1+\left(2-1\right)\cdot2\cdot\left(2+1\right)+2+...+\left(100-1\right)\cdot100\cdot\left(100+1\right)+100\)
\(=1+2+1\cdot2\cdot3+...+99\cdot100\cdot101\)
\(=5050+25497450\)
\(=25502500\)
24 tháng 11 2021
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16 tháng 4 2017
a)Ta có :
\(A=\dfrac{1}{4}+\dfrac{1}{4^2}+\dfrac{1}{4^3}+............+\dfrac{1}{4^{100}}\)
\(4A=1+\dfrac{1}{4}+\dfrac{1}{4^2}+\dfrac{1}{4^3}+..........+\dfrac{1}{4^{99}}\)
\(4A-A=\left(1+\dfrac{1}{4}+.......+\dfrac{1}{4^{99}}\right)-\left(\dfrac{1}{4}+\dfrac{1}{4^2}+.....+\dfrac{1}{4^{100}}\right)\)
\(3A=1-\dfrac{1}{4^{100}}\)
\(\Rightarrow A=\dfrac{1-\dfrac{1}{4^{100}}}{3}\)
~ Chúc bn học tốt ~
HT
1
3 tháng 10 2018
Ta có:
\(100-\left(1+\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{100}\right)\)
\(=\left(1-1\right)+\left(1-\dfrac{1}{2}\right)+\left(1-\dfrac{1}{3}\right)+...+\left(1-\dfrac{1}{100}\right)\)
\(=0+\dfrac{1}{2}+\dfrac{2}{3}+...+\dfrac{99}{100}\)
\(=\dfrac{1}{2}+\dfrac{2}{3}+\dfrac{3}{4}+...+\dfrac{99}{100}\)
3xB=3x(1+3+3^2+3^3+...+3^100)
3xB=3+3^2+3^3+3^4+...+3^100+3^101
3xB-B=3+3^2+3^3+3^4+...+3^100+3^101-1-3-3^2-3^3-...-3^100
2B=3^101-1
B=(3^101-1):2