Question

exercitiul 5 testul 6
varoggg​

Answer 1

 

$\displaystyle\\\text{Testul 6}\\\text{Ex. 5}\\\\S=1+3+3^2+3^3+...+3^{80}\\\\S=3^0+3^1+3^2+3^3+...+3^{80}\\\\a)\\\\\boxed{\Big(S:13=c\^at~~\text{rest}~0\Big)\text{ este echivalent cu }\Big(S~\vdots~13\Big)}\\\\\text{Exponentii 0 la 80}\\\\\implies~\boxed{\bf~n=81~de~termeni}$

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$\displaystyle\\\text{Observam ca suma primilor 3 termeni este }~3^0+3^1+3^2=1+3+9=13\\\\\text{Vom grupa pe cei 81 de termeni in grupe de cate 3 termeni.}\\\\S=(1+3+3^2)+(3^3+3^4+3^5)+...+(3^{78}+3^{79}+3^{80})\\\\S=(1+3+3^2)+3^3(1+3+3^2)+...+3^{78}(1+3+3^2)\\\\S= (1+3+3^2)(1+3^3+...+3^{78})\\\\S=13(1+3^3+...+3^{78})\\\\\implies~\boxed{\bf~S~\vdots~13~~\text{\bf deoarece un termen = 13.}}$

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$\displaystyle\\ b)\\\\\text{Calculam suma S}\\\\S=3^0+3^1+3^2+3^3+...+3^{80}\\\\S=\frac{3^{80+1}-1}{3-1}\\\\S=\frac{3^{81}-1}{2}\\\\2S=2\times \frac{3^{81}-1}{2}\\\\2S=3^{81}-1\\\\2S=3^{3\times27}-1\\\\2S=\Big(3^3\Big)^{27}-1\\\\2S=27^{27}-1\\\\27^{27}-1>27^{27}-9\\\text{Deoarece }~27^{27}=27^{27}\\\text{dar din stanga egalului se scade doar 1 iar din dreapta se scade 9.}\\\\\text{de exemplu:}~(10-1)>(10-9)\\\\\implies~\boxed{\bf~2S>27^{27}-9}$

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