Déterminons dans chacun des cas la valeur de n:

a. On a :

[tex](\frac{4^{5}+4^{5}+4^{5}+4^{5} }{3^{5}+3^{5}+3^{5} } )(\frac{6^{5}+6^{5}+6^{5}+6^{5}+6^{5}+6^{5} }{2^{5}+2^{5} } )= (\frac{4^{5}\times4 }{3^{5}\times3 } )(\frac{6^{5}\times6 }{2^{5}\times2 } )\\(\frac{4^{5}+4^{5}+4^{5}+4^{5} }{3^{5}+3^{5}+3^{5} } )(\frac{6^{5}+6^{5}+6^{5}+6^{5}+6^{5}+6^{5} }{2^{5}+2^{5} } )= \frac{4^{6} }{3^{6} } \times\frac{6^{6} }{2^{6} } \\[/tex]

[tex](\frac{4^{5}+4^{5}+4^{5}+4^{5} }{3^{5}+3^{5}+3^{5} } )(\frac{6^{5}+6^{5}+6^{5}+6^{5}+6^{5}+6^{5} }{2^{5}+2^{5} } )=\frac{4^{6} }{3^{6} } \times(\frac{6}{2} )^{6} \\(\frac{4^{5}+4^{5}+4^{5}+4^{5} }{3^{5}+3^{5}+3^{5} } )(\frac{6^{5}+6^{5}+6^{5}+6^{5}+6^{5}+6^{5} }{2^{5}+2^{5} } )=\frac{4^{6} }{3^{6} } \times3^{6} \\(\frac{4^{5}+4^{5}+4^{5}+4^{5} }{3^{5}+3^{5}+3^{5} } )(\frac{6^{5}+6^{5}+6^{5}+6^{5}+6^{5}+6^{5} }{2^{5}+2^{5} } )= 4^{6} \\[/tex]

[tex](\frac{4^{5}+4^{5}+4^{5}+4^{5} }{3^{5}+3^{5}+3^{5} } )(\frac{6^{5}+6^{5}+6^{5}+6^{5}+6^{5}+6^{5} }{2^{5}+2^{5} } )= (2^{2}) ^{6} \\(\frac{4^{5}+4^{5}+4^{5}+4^{5} }{3^{5}+3^{5}+3^{5} } )(\frac{6^{5}+6^{5}+6^{5}+6^{5}+6^{5}+6^{5} }{2^{5}+2^{5} } )= 2^{12}[/tex]

et On a

[tex](\frac{4^{5}+4^{5}+4^{5}+4^{5} }{3^{5}+3^{5}+3^{5} } )(\frac{6^{5}+6^{5}+6^{5}+6^{5}+6^{5}+6^{5} }{2^{5}+2^{5} } ) = 2^{n}[/tex]

Alors [tex]2^{12}= 2^{n} \\[/tex]

Donc n = 12

b.  On a :

[tex]3^{2001} +3^{2002} +3^{2003} = 3^{2001} +3\times3^{2001} +3^{2} \times3^{2001}\\3^{2001} +3^{2002} +3^{2003} = 3^{2001} (1+3+3^{2} )\\3^{2001} +3^{2002} +3^{2003} = 3^{2001} (4+9 )\\3^{2001} +3^{2002} +3^{2003} = 13\times3^{2001}[/tex]

et on a

[tex]3^{2001} +3^{2002} +3^{2003} = n3^{2001}[/tex]

Donc n = 13

c. On a

[tex](10^{2002}+25 )^{2} - (10^{2002}-25)^{2} = 10^{2004}+50\times10^{2002}+5^{4} - 10^{2004}+50\times10^{2002}-5^{4}= 2\times50\times10^{2002} = 100\times10^{2002}= 10^{2} \times10^{2002}=10^{2004}[/tex]

et on a

[tex](10^{2002}+25 )^{2} - (10^{2002}-25)^{2} = 10^{n}[/tex]

Alors [tex]10^{2004} = 10^{n}[/tex]

Donc n = 2004