giup toi vs nha
N=1+3^1+3^2+…+3^99
Biet 1^2+2^2+3^2+…+10^2=385
Tinh2^2+4^2+…+20^2
Thank you
giup toi vs nha N=1+3^1+3^2+…+3^99 Biet 1^2+2^2+3^2+…+10^2=385 Tinh2^2+4^2+…+20^2 Thank you
By Aaliyah
By Aaliyah
giup toi vs nha
N=1+3^1+3^2+…+3^99
Biet 1^2+2^2+3^2+…+10^2=385
Tinh2^2+4^2+…+20^2
Thank you
Bạn tham khảo :
$N = 1+3^1+3^2+..+3^{99}$
$3N = 3+3^2+3^3+..+3^{100}$
$3N – N =( 3+3^2+3^3+..+3^{100})-( 1+3^1+3^2+..+3^{99})$
$2N = 3^{100} – 1$
$N = 3^{100} – 1:2$
$2^2 +4^2+6^2+……….+20^2$
$= (1.2)^2+(2.2)^2+(3.2)^2+…+(10.2)^2$
$= 1^2.2^2+2^2.2^2+3^2.2^2+…10^2.2^2$
$= (1^2+2^2+ 3^2+………..10^2).2^2$
$= 385.4$
$= 1540$
Giải thích các bước giải:
\(\begin{array}{l}
N = 1 + {3^1} + {3^2} + {3^3} + …. + {3^{99}}\\
\Rightarrow 3N = 3 + {3^2} + {3^3} + {3^4} + …. + {3^{100}}\\
\Rightarrow 3N – N = \left( {3 + {3^2} + {3^3} + {3^4} + …. + {3^{100}}} \right) – \left( {1 + 3 + {3^2} + {3^3} + …. + {3^{99}}} \right)\\
\Leftrightarrow 2N = {3^{100}} – 1\\
\Leftrightarrow N = \frac{{{3^{100}} – 1}}{2}
\end{array}\)
Lại có:
\(\begin{array}{l}
{2^2} + {4^2} + {6^2} + …. + {20^2}\\
= {\left( {2.1} \right)^2} + {\left( {2.2} \right)^2} + {\left( {2.3} \right)^2} + …. + {\left( {2.10} \right)^2}\\
= {2^2}{.1^2} + {2^2}{.2^2} + {2^2}{.3^2} + …. + {2^2}{.10^2}\\
= {2^2}.\left( {{1^2} + {2^2} + {3^2} + …. + {{10}^2}} \right)\\
= 4.385 = 1540
\end{array}\)