Respuesta :
Answer: The proof is done below.
Step-by-step explanation: We are given to prove that the following statement is true :
"For every natural number n, [tex]n^5+4n[/tex] is a multiple of 5."
We will prove the above statement by MATHEMATICAL INDUCTION.
Let n = 1. Then, we get
[tex]n^5+4n=1^5+4\times5=5,[/tex] a multiple of 5.
Let n = 2. Then, we get
[tex]n^5+4n=2^5+4\times2=40,[/tex] a multiple of 5.
Let the statement be true for n = m, where m is a natural number.
So,
[tex]m^5+4m=5k,[/tex] for any natural number k.
Then,
[tex](m+1)^5+4(m+1)\\\\=m^5+5m^4+10m^3+10m^2+5m+1+4m+4\\\\=(m^5+4m)+5m^4+10m^3+10m^2+5m+5\\\\=5k+5(m^4+2m^3+2m^2+m+1)\\\\=5(k+m^4+2m^3+2m^2+m+1),[/tex] which is a multiple of 5.
Therefore, if the statement is true for n = m, then it is true for n = m+1.
That is, the statement is true for all natural numbers n.
Hence proved.
