Question

Primele 3 sub puncte,va rog!​Dau coroana!

Answer 1

 

a)

$\displaystyle\bf\\2(3+3^2+3^3+.~.~.+3^{2000})=2\left(\frac{3^{2001}-1}{3-1}-1\right)=\\\\=3^{2001}-1-2=3^{2001}-3\\\\3^{2001}-3=4\cdot9^{998}\cdot x-3^{2000}-3~~~\Big|+3\\\\3^{2001}=4\cdot9^{998}\cdot x-3^{2000}\\\\3^{2001}=4\cdot\Big(3^2\Big)^{998}\cdot x-3^{2000}\\\\3^{2001}=4\cdot3^{2\cdot998}\cdot x-3^{2000}\\\\3^{2001}=4\cdot3^{1996}\cdot x-3^{2000}\\\\3^{1996+5}=4\cdot3^{1996}\cdot x-3^{1996+4}\\\\3^{1996}\cdot3^5=4\cdot3^{1996}\cdot x-3^{1996}\cdot 3^4~~~\Big|:3^{1996}\\\\3^5=4x-3^4$

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$\displaystyle\bf3^5=4x-3^4\\4x=3^5+3^4\\4x=243+81\\4x=324\\\\x=\frac{324}{4}\\\\\boxed{\bf x=81}$

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b)

$\displaystyle\bf\\3(4+4^2+4^3+.~.~.+4^{2000})=5\cdot1024^{399}\cdot x-4^{2000}-4\\\\3\left(\frac{4^{2001}-1}{4-1}-1\right)=5\cdot\Big(4^5\Big)^{399}\cdot x-4^{2000}-4\\\\4^{2001}-1-3=5\cdot4^{1995}\cdot x-4^{2000}-4\\\\4^{1995+6}-4=5\cdot4^{1995}\cdot x-4^{1995+5}-4~~~\Big|+4\\\\4^{1995}\cdot4^6=5\cdot4^{1995}\cdot x-4^{1995}\cdot4^5~~~\Big|:4^{1995}\\\\4^6=5x-4^5\\5x=4^6+4^5\\5x=4^5(4+1)\\5x=4^5\cdot5~~~\Big|:5\\\boxed{\bf x=4^5}$

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c)

$\displaystyle\bf\\2+2\cdot3+2\cdot3^2+2\cdot3^3+.~.~.+2\cdot3^{75}=9^{2x}-1\\\\2(1+3+3^2+3^3+.~.~.+3^{75})=\Big(3^2\Big)^{2x}-1\\\\2\left(\frac{3^{76}-1}{3-1}\right)=3^{\b2\b\cdot\b2\b x}-1\\\\3^{76}-1=3^{4x}-1~~~\Big|+1\\\\3^{76}=3^{4x}\\\\4x=76\\\\x=\frac{76}{4}\\\\\boxed{\bf x=19}$