Question #275904

Prove that 3n+4n+5n is divisible by 12 whenever n is an odd positive integer.


1
Expert's answer
2021-12-06T16:26:38-0500

As i understand it should be 3n+4n+5n3^n+4^n+5^n

To prove that number is divisible by 12 we will prove that it is divisible by both 3 and 4


Divisible by 3:

3n3^n is obviously divisible by 3 for every odd integer n, then the point is to prove that 4n+5n4^n+5^n is divisible by 3

According to abbreviated multiplication formula for odd degrees:

4n+5n=(4+5)(4n14n25+4n352...45n2+5n1)=9(4n14n25+4n352...45n2+5n1)34^n+5^n=(4+5)(4^{n-1}-4^{n-2}*5+4^{n-3}*5^2-...-4*5^{n-2}+5^{n-1})=9*(4^{n-1}-4^{n-2}*5+4^{n-3}*5^2-...-4*5^{n-2}+5^{n-1})⋮3

Divisible by 4:

4n4^n is obviously divisible by 4 for every odd integer n, then the point is to prove that 3n+5n3^n+5^n is divisible by 4

According to abbreviated multiplication formula for odd degrees:

3n+5n=(3+5)(3n13n25+3n352...35n2+5n1)=8(3n13n25+3n352...35n2+5n1)43^n+5^n=(3+5)(3^{n-1}-3^{n-2}*5+3^{n-3}*5^2-...-3*5^{n-2}+5^{n-1})=8*(3^{n-1}-3^{n-2}*5+3^{n-3}*5^2-...-3*5^{n-2}+5^{n-1})⋮4


The statement has been proven


Need a fast expert's response?

Submit order

and get a quick answer at the best price

for any assignment or question with DETAILED EXPLANATIONS!

Comments

No comments. Be the first!
LATEST TUTORIALS
APPROVED BY CLIENTS