Question

Sa se arate ca numarul A=5la n X 4la n+3 + 5la n+1 X 4 la n+2 + 5la n +3 X 4 la n - 5la n + 2 X 4 la n+1este divizibil cu 13 pt orice n apartine la N
X=semnul de inmultire 
sa se arate ca nr A= 3 la n X 7 la n+2 - 3la n+1 X  7la n+1 + 3 la n+2 X 7la n+1 este divizibil cu 91 pt orice n apartine la N
mersi :*

Answer 1

$A= 5^{n}*4 ^{n+3}+ 5^{n+1}*4 ^{n+2}+5 ^{n+3} *4 ^{n}-5 ^{n+2} *4 ^{n+1}= \\ =5^{n}*4 ^{n}*(4 ^{3}+5 ^{1}*4 ^{2}+5 ^{3}-5 ^{2}*4 ^{1})= \\ = 5^{n}*4 ^{n}*(64+80+125-100)= \\ =5^{n}*4 ^{n}*169= \\ =5^{n}*4 ^{n}*13 ^{2} \\ A= 3^{n}*7 ^{n+2}- 3^{n+1}*7 ^{n+1}+3 ^{n+2} *7 ^{n+1}= \\ =3^{n}* 7^{n}*(7 ^{2} -3*7+3 ^{2}*7) = \\ =3^{n}* 7^{n}*91$