The first formula is 2

The second formula is 3*(2

I claim that these two formulas are equivalent.

It may look counterintuitive that if i is even then 2

Below I provide a proof using mathematical induction to show off my mad skillz.

Proof by weak mathematical induction:

3*(2

Since 2

Replacing i with i+2 yields

2

2

2

2

Similarly, suppose that 2

Replacing i with i+2 yields

2

2

2

2

.'.

^{i}+ (2^{i-1}- 3) if i is even or 2^{i}+ (2^{i-1}+ 3) if i is odd for all i > 2The second formula is 3*(2

^{i-1}- 1) if i is even or 3*(2^{i-1 }+ 1) if i is odd for all i > 2I claim that these two formulas are equivalent.

It may look counterintuitive that if i is even then 2

^{i}+ (2^{i-1}- 3) = 3*(2^{i-1}- 1) and that if i is odd then 2^{i}+ (2^{i-1}+ 3) = 3*(2^{i-1}+ 1) but I can actually prove it.Below I provide a proof using mathematical induction to show off my mad skillz.

**Claim:**For all i > 2, if i is even then 2^{i}+ (2^{i-1}- 3) = 3*(2^{i-1}- 1) and if i is odd then 2^{i}+ (2^{i-1}+ 3) = 3*(2^{i-1}+ 1).Proof by weak mathematical induction:

__Base Case: i = 3____2__

^{3}+ (2^{3-1}+ 3) = 8 + (4 + 3) = 153*(2

^{3-1}+ 1) = 3*(4+1) = 15 Clearly, base case holds__Inductive Step:____Suppose that 2__

^{i}+ (2^{i-1}+ 3) = 3*(2^{i-1}+ 1) is true for an odd i, must show that the claim is also true for the next odd i, namely i + 2.Since 2

^{i}+ (2^{i-1}+ 3) = 3*(2^{i-1}+ 1) then 2^{i}= 3*(2^{i-1}+ 1) - (2^{i-1}+ 3)Replacing i with i+2 yields

2

^{i + 2}= 3*(2^{i +2 - 1}+ 1) - ( 2^{i + 2 - 1}+ 3)2

^{i + 2}= 3*2^{i+1 }+ 3 - 2^{i+1 }- 32

^{i + 2}= 3*2^{i+1 }+ 0 - 2^{i+1}2

^{i + 2}^{ }= 2*2^{i+1 }= 2^{i + 2}Similarly, suppose that 2

^{i}+ (2^{i-1}- 3) = 3*(2^{i-1}- 1) is true for an even i, must show that the claim is also true for the next even i, namely i + 2Replacing i with i+2 yields

2

^{i + 2}= 3*(2^{i +2 - 1}- 1) - ( 2^{i + 2 - 1}- 3)2

^{i + 2}= 3*2^{i+1 }- 3 - 2^{i+1 }+ 32

^{i + 2}= 3*2^{i+1 }- 0 - 2^{i+1}2

^{i + 2}^{ }= 2*2^{i+1 }= 2^{i + 2}.'.