\(S=4+4^2+4^3+...+4^{101}=?\)
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2 tháng 9 2020
A=1/3+2/3^2+3/3^3+4/3^4+...+101/3^101
3A=1+2/3^1+3/3^2+4/3^3+...+101/3^100
3A-A=(1+2/3^1+3/3^2+4/3^3+...+101/3^100)-(1/3+2/3^2+3/3^3+4/3^4+...+101/3^101)
2A=(1-101/3^101)+(2/3-1/3)+(3/3^2-2/3^2)+...+(101/3^100-99/3^100)
2A=1-101/3^101+(1/3+1/3^2+,,,+1/3^100)
(1/3+1/3^2+,,,+1/3^100) đặt B
B=1/3+1/3^2+,,,+1/3^100
3B=1+1/3+...+1/3^99
3B-B=(1+1/3+...+1/3^99)-(1/3+1/3^2+,,,+1/3^100)
2B=1-1/3^100
B=(1-1/3^100):2=1/2-1/3^100.2
thay B vào 2A ta đc
2A=1-101/3^101+B
2A=1-101/3^101+1/2-1/3^100.2
2A=(1+1/2)-(101/3^101+1/3^100.2)
2A=3/2-(101/3^101+1/3^100.2)
A=3/4-(101/3^101+1/3^100.2)<3/4
TH
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AK
19 tháng 7 2018
\(S=1.2+2.3+3.4+...+101.102\)
\(\Leftrightarrow3S=1.2.3+2.3.3+3.4.3+...+101.102.3\)
\(\Leftrightarrow3S=1.2.3+2.3.\left(4-1\right)+3.4.\left(5-2\right)+...+101.102.\left(103-100\right)\)
\(\Leftrightarrow3S=1.2.3+2.3.4-1.2.3+3.4.5-2.3.4+...+101.102.103-100.101.102\)
\(\Leftrightarrow3S=\left(1.2.3-1.2.3\right)+\left(2.3.4-2.3.4\right)+...+\left(100.101.102-100.101.102\right)\)
\(+101.102.103\)
\(\Leftrightarrow3S=101.102.103\)
\(\Leftrightarrow S=\frac{101.102.103}{3}\)
\(\Leftrightarrow S=353702\)
Vậy \(S=353702\)
14 tháng 12 2021
Sửa: \(S=1+\left(-2\right)+3+\left(-4\right)+...+99+\left(-100\right)+101\)
\(S=-1-1-1-...-1+101\)
S có \(\left(100-1+1\right):2=50\) số hạng \(-1\)
Do đó \(S=\left(-1\right)\cdot50+101=-50+101=51\)
Ta có:
S=4+42+43+...+4101
\(\Leftrightarrow\)4S=42+43+...+4101+4102
\(\Leftrightarrow\)3S=4102-4
\(\Leftrightarrow\)S=\(\frac{4^{102}-4}{3}\)
Vậy S=\(\frac{4^{102}-4}{3}\)
\(S=4+4^2+4^3+...+4^{101}\)
\(4S=4^2+4^3+...+4^{102}\)
\(4S-S=\left(4^2+4^3+...+4^{102}\right)-\left(4+4^2+...+4^{101}\right)\)
\(3S=4^2+4^3+...+4^{102}-4-4^2-...-4^{101}\)
\(3S=\left(4^2-4^2\right)+\left(4^3-4^3\right)+...+\left(4^{101}-4^{101}\right)+\left(4^{102}-1\right)\)
\(3S=0+0+...+0+4^{102}-1\)
\(S=\frac{4^{102}-1}{3}\)
Vậy \(S=\frac{4^{102}-1}{3}\)
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