We are looking for Ax:n or E[v^(K+1)]
E[v^(K+1)]=v^(k+1) * P(K=k)
P(K=0) is the probability of Henry dying in the first year which is q30=(1-e^-u)
P(K=1) is the probability of Henry surviving the first year and dying the second
p30*q31=(e^-u)*(1-e^-u)
With constant force, qx is the same for all x.
P(K=2) is the probability of Henry surviving the first two years. He will
receive payment the third year
regardless of whether he lives or dies in the third year. 2p30=e-2u
sum (over k) P(K=k)=1-x+x-y+y=1
where x=e^-u and y=e^-2u
Hope this helps.