Answer:
Step-by-step explanation:
To prove that [tex]8^n-1[/tex] is a multiple of 7.
Proof by induction:
Let n =1. Then we have 8-1 =7 is a multiple of 7
Thus the P(1) is true.
Let us assume that P(n) is true.
[tex]8^n-1[/tex] = 7l is true for some integer l
To check for P(n+1)
[tex]8^{n+1} -1[/tex]=LHS
=[tex]8^n(8)-1\\=(7l+1)8-1\\=56l+7\\=7(8l+1)[/tex]
Thus we find that this is also a multiple of 7.
If true for n, then true for n+1
Hence we find that since true for 1, we have it is true for all natural numbers.
Answer: with Step-by-step explanation:
We are given that
P(n)=[tex]8^n-1[/tex]
We have to prove that given statement is a multiple of 7 using mathematical induction for all natural numbers n belongs to N.
Suppose n=1
Then P(1)=8-1=7
Hence, it is a multiple of 7 .Therefore, it is true for n=1
We suppose that it is true for n=k
Then P(k)=[tex]8^k-1[/tex] is a multiple of 7.
We shall prove that it is true for n=k+1
P(k+1)=[tex]8^{k+1}-1[/tex] is a multiple of 7
LHS=[tex]8^{k+1}-1[/tex]
=[tex]8^k\cdot8-1[/tex]
=[tex]8^k\cdot8-8+8-1[/tex]
=[tex]8(8^k-1)+7[/tex]
=[tex]8\cdot 7a+7[/tex] because [tex]8^k-1[/tex] is a multiple of 7 therefore[tex]8^k-1=7a[/tex]
=[tex]7(8a+1)[/tex]
P(k+1) is a multiple of 7.
Therefore, P(n) is true for all natural numbers belongs to N.
