Prove by induction that 5^n - 1 is an integer multiple of 2

Question

Prove by induction that 5^n - 1 is an integer multiple of 2

Answer 1

Show that the equation holds when n = 1: when n = 1, 5ⁿ - 1 = 5 - 1 = 4, which is a multiple of 2.

Assume that 5^k - 1 is a multiple of 2 and show that 5^k+1 - 1 is a multiple of 2:

Since 5^k - 1 is a multiple of 2, 5^k - 1 = 2m, for some integer m.

So, 5^k+1 - 1 = 5(5^k) - 1 = 5(5^k) + 4 - 5 = 5(5^k) - 5 + 4 = 5(5^k - 1) + 4 = 5(2m) + 2(2) = 2(5m + 2)

Therefore, 5^k+1 - 1 is a multiple of 2.

By induction, 5ⁿ - 1 is a multiple of 2 for all positive integers, n.

Answer 2

5 raised to any integer power will be a number ending in the digit 5 which is an odd number

subtract 1 from any odd number and you get an even number. All even numbers are divisible evenly by 2.

It's nice to know what you're trying to prove is actually true.

2 divides 5^n -1

2(5^n-1)=v

5^n-1=v/2

5^n=v/2+1

5=(v/2 +1)^(1/n)= nth root of (v/2 +1)

let n=1, then 2(5-1)=2(4)=8 = an even number

let n=2 then 2(25-1)=2(24)=28 = an even number

assume n=k, then k(25-1) = 24k = an even number

try to show if n=k-1, then 24(k+1) = 24k +24 = an even number

24k is even, as you assumed it was even, and an even number + an even number (like 24) is also even

QED

for proof by mathematical inductions

first two steps, try n=1 and n=2, show they work

3rd step assume n=k works

last step, try to show n=k+1 works