A coin is tossed (m+n) times with m>n. Show that the probability of getting m consecutive heads is `(n+2)/2^(m+1)`

Let H and T denote turning up of the head and tail and X denoter the turningh of head or tail. Then
`P(H)=P(T)=1/2and P(X)=1`
P(HH… m times) (XXX… n times)
`=1/2xx1/2...,m "times" =(1)/(2^(m))`
P[T:H H H… m times} {XX… n-1 times}]
= P (T) P(HH.... m times ). P (XXX... (n-2) times) `=(1)/(2^(m+1))`
If the dequence of heads starts with `(r=1)th` throw, then the first 1-1 throws may be head or tail but rth throw must be tail and we have
P[(XX...(r-1)times) t (HH...m times)`xx`
`" "(XX...(n-m-r)"times")xx`
`(1)/(2^(m+1))`
Since all the above cases are matually exclusive, the required prabability is
`(1)/(2^(m))+[(1)/(2^(m+1))+(1)/(2^(m+1))...+ n "times"]=(1)/2^(m)+(n)/(2^(m+1))=(n+2)/(2^(m+1))`