Q.How many op-amps are required to implement this equation? $V_o = -\left(\dfrac{R_f}{R_1}V_1 + \dfrac{R_f}{R_2}V_2 + \dfrac{R_f}{R_3}V_3\right)$

The summing amplifier, or inverting adder, extends the inverting amplifier to several inputs. Each input $V_1, V_2, V_3, \dots$ is applied through its own resistor $R_1, R_2, R_3, \dots$ to the common inverting $(-)$ node, which is a virtual ground; a single feedback resistor $R_f$ sets the overall scale, and the $(+)$ terminal is grounded.

Because the inverting node sits at $0\ \text{V}$ and draws no current, each input drives its own current $V_n/R_n$ into the node; these currents add and flow together through $R_f$. The result is a negative weighted sum:

Each input's weight is $R_f/R_n$, so choosing the resistors sets each coefficient independently. If all resistors are equal, $V_o=-(V_1+V_2+V_3)$, a perfect adder. Only one op-amp is needed for any number of inputs. …