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

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1. Bạn tham khảo :

   $N = 1+3^1+3^2+..+3^{99}$

   $3N = 3+3^2+3^3+..+3^{100}$

    $3N – N  =( 3+3^2+3^3+..+3^{100})-( 1+3^1+3^2+..+3^{99})$

   $2N =     3^{100} – 1$

   $N = 3^{100} – 1:2$ 

   $2^2 +4^2+6^2+……….+20^2$

   $= (1.2)^2+(2.2)^2+(3.2)^2+…+(10.2)^2$

   $= 1^2.2^2+2^2.2^2+3^2.2^2+…10^2.2^2$

   $= (1^2+2^2+ 3^2+………..10^2).2^2$

   $= 385.4$

   $= 1540$

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2. 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:

   \(\begin{array}{l}
   {2^2} + {4^2} + {6^2} + …. + {20^2}\\
    = {\left( {2.1} \right)^2} + {\left( {2.2} \right)^2} + {\left( {2.3} \right)^2} + …. + {\left( {2.10} \right)^2}\\
    = {2^2}{.1^2} + {2^2}{.2^2} + {2^2}{.3^2} + …. + {2^2}{.10^2}\\
    = {2^2}.\left( {{1^2} + {2^2} + {3^2} + …. + {{10}^2}} \right)\\
    = 4.385 = 1540
   \end{array}\)

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