Question

calculati
$[( \sqrt{3} ^{-1}) *( \sqrt{3}^{-2}) *( \sqrt{3}^{-3})*...*( \sqrt{3}^{-100})$ * $3^{2525}$

Answer 1

[(√3⁻¹)*(√3⁻²)*(√3⁻³)*...*(√3⁻¹⁰⁰)]*3²⁵²⁵=

$=[\frac{1}{ \sqrt{3}} * \frac{1}{ \sqrt{3}^2} * \frac{1}{ \sqrt{3}^3} *...* \frac{1}{ \sqrt{3}^{100}} ]*3^{2525}=$      (&)

Observam ca in paranteza, la numitor avem :
√3*√3² *√3³ *√3⁴ *√3⁵ *√3⁶ *√3⁷ *√3⁸ *...*√3¹⁰⁰=
ii grupam pe cei pari si pe cei impari=>
=(√3² *√3⁴ *√3⁶*√3⁸ *...*√3¹⁰⁰)*(√3*√3³ * √3⁵ *√3⁷  * √3⁹ *√3¹¹*...*√3⁹⁷ *√3⁹⁹)=
=(3¹*3²*3³*3⁴*...*3⁵⁰)*(3²*3⁶*3¹⁰*...*3⁹⁸)=
=(3¹⁺²⁺³⁺⁴⁺ ⁻⁻⁻ ⁺⁵⁰) * (3²⁺⁶⁺¹⁰⁺ ⁻⁻⁻ ⁺⁹⁸) =3 ¹²⁷⁵ * 3¹²⁵⁰=3²⁵²⁵
caci 1+2+3+...+50=50*51:2=1275
2+6+10+...+98=(98+2)*[(98-2):4+1]:2=1250


(&)=> $\frac{1}{ 3^{2525} } * 3^{2525} = 1$