Induction (1 Viewer)

[phoenix159]

Prove that 2ⁿ⁺² + 3²ⁿ⁺¹ is divisible by 7 for all integers, n > 0

So far I have:

Let n = 1

2³ + 3³ = 8 + 27 = 35, which is divisible by 7

Assume true for n = k

2^k+2 + 3^2k+1 = 7M

Consider n = k + 1

2^k+3 + 3^2k+3
= 2^k.2³ + 3^2k + 3³
= 2^k.8 + 9^k.27
= 2^k.8 + 9.(3.9^k)
= 2^k.8 + 9.(7M - 2.2^k)
= 2^k.8 + 63M - 18.2^k
= 2^k.8 + 63M - 18.2^k
= 63M - 10.2^k
which is not divisible by 7

 

[Kurosaki]

Prove that 2ⁿ⁺² + 3²ⁿ⁺¹ is divisible by 7 for all integers, n > 0

So far I have:

Let n = 1

2³ + 3³ = 8 + 27 = 35, which is divisible by 7

Assume true for n = k

2^k+2 + 3^2k+1 = 7M

Consider n = k + 1

2^k+3 + 3^2k+3
= 2^k.2³ + 3^2k + 3³
= 2^k.8 + 9^k.27
= 2^k.8 + 9.(3.9^k)
= 2^k.8 + 9.(7M - 2.2^k)
= 2^k.8 + 63M - 18.2^k
= 2^k.8 + 63M - 18.2^k
= 63M - 10.2^k
which is not divisible by 7

What you want to do is find from your assumption step. the rest should fall into place once you multiply by 9 and then substitute that into your n=k+1 step.

 

[Trebla]

Prove that 2ⁿ⁺² + 3²ⁿ⁺¹ is divisible by 7 for all integers, n > 0

So far I have:

Let n = 1

2³ + 3³ = 8 + 27 = 35, which is divisible by 7

Assume true for n = k

2^k+2 + 3^2k+1 = 7M

Consider n = k + 1

2^k+3 + 3^2k+3
= 2^k.2³ + 3^2k + 3³
= 2^k.8 + 9^k.27
= 2^k.8 + 9.(3.9^k)
= 2^k.8 + 9.(7M - 2.2^k)
= 2^k.8 + 63M - 18.2^k
= 2^k.8 + 63M - 18.2^k
= 63M - 10.2^k
which is not divisible by 7

2^k+3 + 3^2k+3
= 2.2^k+2 + 9.3^2k+1
= 2.2^k+2 + 9.3^2k+1
= 2.2^k+2 + 9(7M - 2^k+2)
= 63M - 7.2^k+2
= 7(9M - 2^k+2)

 

[phoenix159]

Thanks!