So sanh A,B
A=1+ $3^{1}$+$3^{2}$+ $3^{3}$+…+ $3^{200}$
B= $3^{201}$
So sanh A,B A=1+ $3^{1}$+$3^{2}$+ $3^{3}$+…+ $3^{200}$ B= $3^{201}$
By Lydia
By Lydia
So sanh A,B
A=1+ $3^{1}$+$3^{2}$+ $3^{3}$+…+ $3^{200}$
B= $3^{201}$
=1+31+32+...+3200↔3A=31+32+33...+3201→3A−A=3201−1↔A=3201−12A=1+31+32+…+3200↔3A=31+32+33…+3201→3A−A=3201−1↔A=3201−12
Vì 3201>3201−1>3201−12→B>A3201>3201−1>3201−12→B>A
Đáp án:
B>A
Giải thích các bước giải:
\(\begin{array}{l}
A = 1 + {3^1} + {3^2} + … + {3^{200}}\\
\leftrightarrow 3A = {3^1} + {3^2} + {3^3}… + {3^{201}}\\
\to 3A – A = {3^{201}} – 1\\
\leftrightarrow A = \frac{{{3^{201}} – 1}}{2}
\end{array}\)
Vì \(\begin{array}{l}
{3^{201}} > {3^{201}} – 1 > \frac{{{3^{201}} – 1}}{2}\\
\to B > A\\
\end{array}\)