Question

Cum se rezolva exercitiul urm:{{[(1+2+3+...+100)÷1010] la puterea 2002 ori 5 la puterea 3 ÷5 la puterea 2003} la puterea 3

Answer 1

$1+2+3+...+n= \frac{n(n+1)}{2}$

$[[(1+2+3+...+100):1010] ^{2002} *5 ^{3} :5 ^{2003} ] ^{ 3}$

$=[( \frac{100*101}{2} :1010) ^{2002} *5 ^{3} :5 ^{2003} ] ^{ 3}$

$=(5^{2002} *5 ^{3} :5 ^{2003} ])^{ 3}$

$=(5^{2002+3-2003})^{ 3}$

$=(5 ^{2})^{ 3}$

$=5^{ 2*3}$

$=5^{ 6}$