Question

demonstrati ca nr a= 63 la n +7 la( n+1) * 3 la( 2 n +1) -21 la n *3( n+2 ), n apartine multimea numerelor naturale este divizibil cu 13
demonstrati ca numarul b=35 la n+7 n * 5 la (n+2) +3* 7 la (n +1) *5 la n , n apartine in multimea numerelor naturale este divizibil cu 47
aratati ca numarul A=7*12 la n * 3 la( n+1 ) + 6*4 (n+1) * 9 la(n+2) +18 la(n+1) *2 la (n+1) este divizibil cu 2001 , oricare ar fi n apartina multimea numerelor naturale.

Answer 1

a= $63^{n} + 7^{n+1} * x^{2n+1}- 21^{n} * 3^{n+2}$

 $= 3^{n} * 3^{n} *7^{n} +7^{n}*7*3^{2n}*3-7^{n}*3^{n}*3^{n} * 3^{2} =$

$=7^{n}*3^{2n}*(1+21-9)=$

$7^{n}*3^{2n}*13$


b $=35^{n}+7^{n} * 5^{(n+2)} +3* 7^{(n +1)} *5^{n} =$

$=7^{n}*5^{n}+7^{n} * 5^{n}* 5^{2} +3* 7^{n}*7 *5^{n} =$

$=7^{n}*5^{n}*(1+25+21)= 7^{n}*5^{n}*47$


A $=7*12^{n}*3^{( n+1 )}+6*4^{(n+1)}*9^{(n+2)}+18^{(n+1)}*2^{(n+1)}=$

$=7*12^{n}*3^{n}*3+6*4^{n}*4*9^{n}* 9^{2} +18^{n}*18*2^{n}*2=$

$=7*3^{n}*4^{n}*3^{n}*3+6*4^{n}*4*9^{n}*9^{2}+2^{n}*3^{n}*3^{n}*18*2^{n}*2=$

$=7*9^{n}*4^{n}*3+6*4^{n}*4*9^{n}*9^{2}+4^{n}*9^{n}*18*2=$

$=9^{n}*4^{n}*(7*3+6*4*9^{2}+18*2)=$

$=9^{n}*4^{n}*(21+1944+36)=9^{n}*4^{n}* 2001$