If $\text P(X>k+1)=\cfrac 12 \text P(X>k)$, $k\in \mathbb N^*$ what is the distribution of $X$?
So I did; $\text P(X=k+1)= \text P(X>k)- \text P(X>k+1)=\cfrac 12 \text P(X>k)$ $\text P(X=k)=\cfrac 12 \text P(X>k-1)$ I don't know what to deduce from this.