Let’s assume T(n) is number of assignments of truth value to P$_{1}$, P$_{2}$, … , P$_{n}$ so that
(...(((P$_{1}$ → P$_{2}$) → P$_{3}$) → P$_{4}$) → … → P$_{n-1}$) → P$_{n}$ is true.
Case 1 : P$_{n}$ is true => (...(((P$_{1}$ → P$_{2}$) → P$_{3}$) → P$_{4}$) → … → P$_{n-1}$) → P$_{n}$ is true.
=> P$_{1}$, P$_{2}$, … , P$_{n-1}$ can have any truth value.
=> 2$^{n-1}$ possible assignments.
Case 2 : P$_{n}$ is false => (...(((P$_{1}$ → P$_{2}$) → P$_{3}$) → P$_{4}$) → … → P$_{n-2}$) → P$_{n-1}$ is false.
=> (...(((P$_{1}$ → P$_{2}$) → P$_{3}$) → P$_{4}$) → … → P$_{n-3}$) → P$_{n-2}$ is true and P$_{n-1}$ is false.
=> (...(((P$_{1}$ → P$_{2}$) → P$_{3}$) → P$_{4}$) → … → P$_{n-3}$) → P$_{n-2}$ is true
=> T(n-2) is number of assignments of truth value to P$_{1}$, P$_{2}$, … , P$_{n-2}$ so that
(...(((P$_{1}$ → P$_{2}$) → P$_{3}$) → P$_{4}$) → … → P$_{n-3}$) → P$_{n-2}$ is true.
$_{\therefore }$ T(n) = 2$^{n-1}$ + T(n-2). (recurrence relation)