Eξ= 1/P
As for the specific solution, you need to use extreme knowledge, which is quite complicated:
1. If the meaning of the question ξ obeys the geometric distribution, let q= 1-p, then the (k- 1) power of g(k, p)=q is multiplied by p.
Then Eξ= p+2pq+3 p×q quadratic +4p×q cubic +…+k×p×q (k- 1) power+…
Let Sn = p+2pq+3p× q2+4p× q3+…+n× p× q (n-1).
QSn = pq+2p×Q2+3p×Q3+…+(n- 1)×p×q(n- 1)+n×p×q n。
∴ (1-q) sn = p [1+Q2+Q3+…+q (n-1)]-n×q times n× q.
∴Sn=[ 1-q to the n power ]/[ 1-q]+n×q to the n power.
∴eξ= lim sn =( 1-0)/( 1-q)+n×0 = 1/p
n→∞
I wonder if this friend C2H5-SH can understand it.