Erlang induction step Derive the formula: “Exp(λ)+Erlang(ℓ,λ)∼Erlang(ℓ+1,λ)” Observation: By repeatedly applying the above formula, we see that: “Exp(λ)+⋯+Exp(λ)⏞ℓ terms∼Erlang(ℓ,λ)”