There is no feedback here — every path runs forward from R to C — so the transfer function is simply the sum of the gains of all the forward paths.

Step 1 — work forward to X(S). The first summing junction receives the output of G₁ and a direct connection from R:

\(X(s)=G_{1}R(s)+R(s)=R(s)\left(G_{1}+1\right)\)

Step 2 — through G₂ to the second junction. That junction receives \(G_{2}X(s)\) and again a direct connection from R:

\(C(s)=G_{2}X(s)+R(s)=G_{2}R(s)\left(G_{1}+1\right)+R(s)\)

Step 3 — expand and divide.

\(C(s)=R(s)\left[G_{1}G_{2}+G_{2}+1\right]\)

\(\dfrac{C(s)}{R(s)}=G_{1}G_{2}+G_{2}+1\)

— option 4.

Reading it as three parallel paths gives the same answer and makes the structure plain:

Since the paths are in parallel and no loop exists, the gains simply add — which is Mason's rule with \(\Delta=1\) and every \(\Delta_{k}=1\).

Why option 3 is the trap. It has \(G_{1}\) where the answer needs \(G_{2}\) — that is, it assumes the bypass connection joins after G₂ rather than before it. The decisive question is where each direct line enters: the first bypass enters ahead of G₂, so it is amplified by G₂; the second enters after G₂, so it is not amplified at all. Tracing where a signal enters, rather than where it leaves, is what separates the two options.

A check on the result. Setting \(G_{1}=G_{2}=0\) should leave only the direct connection, and the expression correctly gives 1. Setting \(G_{1}=0\) should leave the second path and the direct one, giving \(G_{2}+1\) — which it does. Option 3 fails this second test, since it would give 1.

The absence of any feedback is why no \(1+GH\) denominator appears: with no loop, the characteristic equation is trivial and the system cannot be made unstable by raising the gain.

Hence, C(s)/R(s) = G₁G₂ + G₂ + 1.