(original question) Prove that the Summation of j=0 to n of (-1/2)^j = (2^{n}^{+1}+ (-1)^{n}) / (3*2^{n})

I have a question that includes the following

P(k) (-1/2)^{0}+ (-1/2)^{1}+ (-1/2)^{2}+ ... + (-1/2)^{k}= (2^{k}^{+1}+ (-1)^{k}) / (3*2^{k})

So I have k+1 as

P(k+1) ((2^{K+1}+ (-1)^{k}) / (3*2^{k})) + (-1/2)^{k+1}= (2^{k}^{+2}+(-1)^{k+1}) / (3*2^{k+1})

But I dont know how to get those sides to be equivalent. Can anyone please help?