need help with this nice math problem

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A coin with p{h}=p-1 is tossed n times. Show that the probability that the number of heads is even equals 0.5[1+(q-p)^n]

I appreciatte your any of your valuable contributions.
Thanks
 

Does p{h} mean p{Heads}? What is q?
 

p-heads
q-probability of not happening hats-(tails)

thanks

Added after 4 hours 55 minutes:

corection to the problem

A coin with p{h}=p=1-q is tossed n times. Show that the probability that the number of heads is even equals 0.5[1+(q-p)^n].
 

I found a hint for this problem. The hint is saying that we should write down the binomial theorem expansions of (q + p)^n and (q – p)^n, then add them together.
This would eventually result in expected solution-- 0.5[1+(q - p)^n]. The only problem is, I can not get to this expected solution. I need you help.
 

That's brilliant! Yes, the sum of all the even terms is just ((q+p)^n + (q-p)^n)/2, which is exactly your answer.
 

where do you get this 1 from? 0.5[1+(q-p)^n]. [/b]
 

ooops!!!
 

p+q sure equals one as it conveys all the probabilities
 

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