Take n=1 6^1+4 =>10 which is divisible by 5.
6^n+4 is divisible by 5 for all nEN.
Take n= n+1
6^n+1 +4 => 6^n*6 +4 => 6[6^n +4]
as 6^n+4 is divisible by 5 for all nEN
6*[multiples of 5] also divisible by 5
Hence 6^n+4 is divisible by 5 for all nEN.
Can anyone please help me with the following question:
Prove by mathematical induction that 6^n + 4 is a multiple of 5, for nEN.
3 answers
mmmmuj
Follow the standard 3-step method
1. show it to be true for n=1
6^1 + 4 = 10 which is divisible by 5
2. Assume it to be true for n = k , that is
6^k + 4 is divisible by 5
3. show that it is also true for n = k+1
that is,
6^(k+1) + 4 is divisible by 5
Use the fact that if 2 numbers are both divisible by some number , then their difference is also divisible by that number. e.g. bot 35 and 14 are divisible by 7, so is 35-14 or 21 divisible by 7
(6^(k+1) + 4) - (6^k + 4)
= 6^(k+1) - 6^k
= 6^k(6^1 - 1)
= 5(6^k)
which is a multiple of 5, thus divisible by 5
Therefore 6^(k+1) + 4 is divisible by 5
1. show it to be true for n=1
6^1 + 4 = 10 which is divisible by 5
2. Assume it to be true for n = k , that is
6^k + 4 is divisible by 5
3. show that it is also true for n = k+1
that is,
6^(k+1) + 4 is divisible by 5
Use the fact that if 2 numbers are both divisible by some number , then their difference is also divisible by that number. e.g. bot 35 and 14 are divisible by 7, so is 35-14 or 21 divisible by 7
(6^(k+1) + 4) - (6^k + 4)
= 6^(k+1) - 6^k
= 6^k(6^1 - 1)
= 5(6^k)
which is a multiple of 5, thus divisible by 5
Therefore 6^(k+1) + 4 is divisible by 5