Question

Prove by the method of induction that:

a.
`\tt x^n-y^n` is divisible by x-y.
b. n(n+1) (n+2) is a multiple of 6.

Answer 1

Explanation:

1)

Let p(n) :
`x^n - y^n` is divisible by (x-y)

For n = 1,

p(1): x¹ - y¹ = x-y

= 1*(x - y)

(x - y) is divisible by (x - y)

∴ p(n) is true for n = 1

Let us assume that p(n) is true for n = k

p(k):
`x^k - y^k` is divisible by (x-y)

i.e.
`x^k - y^k = q(x-y)`

TP : p(n) is true for n = (k + 1)

p(k):
`x^(k+1) - y^(k+1)` is divisible by (x-y)

i.e.
`x^(k+1) - y^(k+1) = z(x-y)`

p(k + 1):

`x^(k+1) - y^(k+1)\\\\= x*x^k-y*y^k\\`

Adding and subtracting
`x^ky`,

`x(x^k)-y(y^k)+x^ky-x^ky\\\\=x(x^k)-x^ky+x^ky-y(y^k)\\\\=x^k(x-y)-y(x^k-y^k)\\\\=x^k(x-y)-yq(x-y)\\\\=(x-y)(x^k-qy)`

Which is divisible by (x-y)

∴ p(k + 1):
`x^(k+1) - y^(k+1)` is divisible by (x-y)

Therefore p(n)
`x^n - y^n` is divisible by (x-y) is true

2)

Let p(n) : n(n+1)(n+2) is a multiple of 6

For n = 1,

p(1) : 1(1 + 1)(1 + 2)

= 1(2)(3)

= 6

which is divisible by 6

∴ p(n) is true for n = 1

Let us assume that p(n) is true for n = k

p(k): k(k+1)(k+2) is a multiple of 6

i.e. k(k+1)(k+2) = 6q

TP : p(n) is true for n = (k + 1)

p(k): (k + 1)(k + 1 + 1)(k + 1 + 2) is a multiple of 6

i.e. (k + 1)(k + 2)(k + 3) = 6z

p(k + 1): (k + 1)(k + 2)(k + 3)

=
`(6q)/(k)(k+3)` (from p(k))

=
`6[(q(k+3))/(k) ]`

= 6z

which is divisible by 6

∴ p(n) is true for n = k

Answer 2

Please look at the explanation!

Explanation:

a. Base case: For n = 1, we have x^1 - y^1 = x - y, which is clearly divisible by x-y.

Inductive step: Assume that for some positive integer k, x^k - y^k is divisible by x-y. We will show that this implies that x^(k+1) - y^(k+1) is also divisible by x-y.

Starting with x^(k+1) - y^(k+1), we can factor out x^k from the first term and y^k from the second term:

x^(k+1) - y^(k+1) = xx^k - yy^k

= x(x^k - y^k) + y^k(x - y)

By our assumption, x^k - y^k is divisible by x-y. Also, x-y is a factor of (x-y), so we can write:

x(x^k - y^k) + y^k(x - y) = (x-y)(x^k - y^k + y^k)

= (x-y)x^k

b. Proof by induction that n(n+1)(n+2) is a multiple of 6.

Base case: For n = 1, we have 1(1+1)(1+2) = 6, which is clearly a multiple of 6.

Inductive step: Assume that for some positive integer k, k(k+1)(k+2) is a multiple of 6. We will show that this implies that (k+1)(k+2)(k+3) is also a multiple of 6.

Starting with (k+1)(k+2)(k+3), we can expand the product:

(k+1)(k+2)(k+3) = k(k+1)(k+2) + 3(k+1)(k+2)

By our assumption, k(k+1)(k+2) is a multiple of 6. Also, 3(k+1)(k+2) is clearly a multiple of 6, since it contains a factor of 3 and a factor of 2. Therefore, (k+1)(k+2)(k+3) is a multiple of 6.