Question

2^3n-1 is dividible by (AnEN)
a) 25
b)8
c)3
d)7
​

Answer 1

P(n):23n−1

If It's divisible by 7 then,

Base case:

n=1 then,

P(n):23n−1=23×1−1=7

If n=k is  true for some natural numbers, then It should also be true for (k+1),

For n=k,

P(k):23k−1=7d  

P(k):23k=7d+1                .....(1)         

Where d=1,2,3,4.......so on

Now we check for n=k+1,

P(k+1):23(k+1)−1=7d  

P(k+1):23k+3=7d+1                       

(Since axy=axay)

7 is the correct answer