Tips on how to solve these proofs by mathematical induction?

Question

I think I understand the basic steps of Mathematical Induction right:

1. Prove the statement for n=1

2. Assume it's true for n=k

3. Prove it's true for n=k+1

But these three proofs seem completely foreign to me:

Sigma 2k-1=n^2

d^n/dx[cosx]=cos[x + n*pi/2]

Prove that 9(^n)-1 is a multiple of 8

If anyone could give me any clues or tips as to how to solve any of these, I'd be very grateful

Answer 1

1.

For n = 1, we have:

Σ [i=1 to 1] (2i-1) = 2(1) - 1 = 1 = 1²

We assume:

Σ [i=1 to k] (2i-1) = k² .

Then:

Σ [i=1 to k+1] (2i-1) =

Σ [i=1 to k] (2i-1) + (2(k+1) - 1) =

k² + (2(k+1) - 1) =

k² + (2k + 2 - 1) =

k² + 2k + 1 =

(k+1)² .

2.

For n = 1, we have:

d¹(cosx)/dx = -sinx = cos(x + 1∙(π/2)).

We assume:

d^k(cosx)/dx = cos(x + k∙(π/2)).

Then:

d^(k+1)(cosx)/dx =

d(d^k(cosx)/dx)/dx =

d(cos(x + k∙(π/2)))/dx =

-sin(x + k∙(π/2)) =

cos((x + k∙(π/2)) + π/2) =

cos(x + (k+1)∙(π/2))

3.

For n = 1, we have:

9¹ - 1 = 8 .

We assume:

9^k - 1 = 8i, for some positive integer i .

Then:

9^(k+1) - 1 =

(9^k)∙(9¹) - 1 =

(9^k - 1 + 1)∙(9) - 1 =

(9^k - 1)∙(9) + (1)∙(9) - 1 =

(8i)∙(9) + (1)∙(9) - 1 =

(8i)∙(9) + 8 =

8(9i) + 8 =

8(9i + 1) =

8j, where j is the positive integer equal to 9i + 1 .

I noticed that there were some extra parentheses floating about.