Answer:
It is true for all values if n
Step-by-step explanation:
Let
[tex]P(n)=5^{n+1}+4.6^{n}-9[/tex]
We have to prove that 20 is a factor of P(n) by mathematical induction.
Let us start:
Step1:
Check P(n) is true for 1.
[tex]P(1)=5^{1+1}+4.6^{1}-9\\P(1)=5^{2}+4.6^{1}-9\\P(1)=25+24-9\\P(1)=40=20*2[/tex]
Hence P(n) is true for n=1
Step 2:
Assume that P(n) is true for some value of n say t
That is P(t) is true or
[tex]5^{t+1}+4.6^{t}-9=20k[/tex]
Where k is some constant…
Step3:
We have to prove that P(t+1) is also true, then P(n) is true for every value of n.
Hence
[tex]P(t+1)= 5^{(t+1+1)}+4*6^{t+1}-9[/tex]
[tex]P(t+1)=5^{t+1}*5+4*6^t*6-9\\=5^{t+1}*5+4*6^t*(5+1)-9\\=5^{t+1}*5+(4*6^t)*5+(4*6^t) -9\\=5(5^{t+1}+4*6^t)+(4*6^t) -9\\=5(20k+9)+ 4*6^t -9\\=100k+45-9+4*6^{t}\\=100k+36+4*6^{t}\\=100k+4(9+*6^{t})\\=100k+4(20k-5^{t+1})\\=100k+80k-4*5*5^t\\=180k-20*5^t\\=20(9k-5^t)\\=20r[/tex]
where r is some constant.
Hence
[tex]P(t+1) [/tex] is also true ,
Thus P(n) is true for all values of n , and we can say that
When [tex]5^{n+1}+4.6^{n}[/tex] is divided by 20, we get 9 as remainder.
