Toán Lớp 7: C1
Cho A=1+3+ $3^{2}$ +…+$3^{100}$
B= $\frac{3^{101} }{2}$
tính B-A
C2
cho A=1+4+$4^{2}$ +…+$4^{99}$
B=$4^{10}$
CMR A< $\frac{B}{3}$
nhanh nha
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Toán Lớp 7: C1
Cho A=1+3+ $3^{2}$ +…+$3^{100}$
B= $\frac{3^{101} }{2}$
tính B-A
C2
cho A=1+4+$4^{2}$ +…+$4^{99}$
B=$4^{10}$
CMR A< $\frac{B}{3}$
nhanh nha
Comments ( 2 )
1)A = 1 + 3 + {3^2} + … + {3^{100}}\\
\Leftrightarrow 3A = 3 + {3^2} + {3^3} + … + {3^{101}}\\
\Leftrightarrow 3A – A = {3^{101}} – 1\\
\Leftrightarrow 2A = {3^{101}} – 1\\
\Leftrightarrow A = \dfrac{{{3^{101}}}}{2} – \dfrac{1}{2}\\
\Leftrightarrow B – A = \dfrac{{{3^{101}}}}{2} – \left( {\dfrac{{{3^{101}}}}{2} – \dfrac{1}{2}} \right) = \dfrac{1}{2}\\
2)A = 1 + 4 + {4^2} + … + {4^{99}}\\
\Leftrightarrow 4A = 4 + {4^2} + {4^3} + … + {4^{100}}\\
\Leftrightarrow 4A – A = 3A = {4^{100}} – 1\\
\Leftrightarrow A = \dfrac{{{4^{101}}}}{3} – \dfrac{1}{3}\\
B = {4^{100}}\\
\dfrac{B}{3} – A = \dfrac{{{4^{100}}}}{3} – \left( {\dfrac{{{4^{100}}}}{3} – \dfrac{1}{3}} \right) = \dfrac{1}{3}\\
\Leftrightarrow A < \dfrac{B}{3}
\end{array}$