math induction help?

Question

Use the Principle of Mathematical Induction to prove that
1 - 2 + 2^2 - 2^3 +...+ (-1)^x * 2^x = [2^(x+1)*(-1)^x + 1] / 3
for all positive integers n

Answer 1

RTP that
∑ (-1)^n * 2^n = [(-1)^n * 2^(n+1) ] / 3

Verify that both RHS = LSH = 1 for n = 0

Now assume that it holds for n = k ie
∑ (-1)^k * 2^k = [(-1)^k * 2^(k+1) ] / 3

We must prove it holds for n = k + 1
∑ to k+1 = ∑ (-1)^k * 2^k + (-1)^(k+1) *2^(k+1)
= [(-1)^k * 2^(k+1) ] / 3 + (-1)^(k+1) *2^(k+1)
= [(-1)^k * 2^(k+1) ] / 3 - (-1)^(k) *2^(k+1) *3 / 3
= [(-1)^k * 2^(k+1) - 3* (-1)^(k) *2^(k+1)] / 3
= [-2 * (-1)^k * 2^(k+1)] / 3
= [(-1)^(k+1) * 2^(k+1+1)] / 3
Proved by induction

Answer 2

First show that it works for x = 0 (the first term)
(-1)^0*2^0 = [2^(0+1)*(-1)^0 +1]/3
1 = (2^1*1+1)/3
1 = 3/3 = 1
It does work for x=0 so let it work for x = k so sum of 1st k terms is: [2^(k+1) *(-1)^k +1]/3
Now the k+1 term is (-1)^(k+1) *2^(k+1)
So the sum of the 1st k+1 terms should be
[2^(k+1) *(-1)^k +1]/3 + (-1)^(k+1) *2^(k+1)
Now show that the above simplifies to [2^(k+1)*(-1)^k+ 1] / 3
and you will have proven the formula by MI.