Prove that ${^{35}}{C}_{5} + \sum_{r = 0}^{4} {^{\left(39 - r\right)}}{C}_{4} = {^{40}}{C}_{5}$
$\begin{aligned} LHS = {^{35}}{C}_{5} + \sum_{r = 0}^{4} {^{\left(39 - r\right)}}{C}_{4} & = {^{35}}{C}_{5} + {^{39}}{C}_{4} + {^{38}}{C}_{4} + {^{37}}{C}_{4} + {^{36}}{C}_{4} + {^{35}}{C}_{4} \\\\ & = \left({^{35}}{C}_{5} + {^{35}}{C}_{4}\right) + {^{36}}{C}_{4} + {^{37}}{C}_{4} + {^{38}}{C}_{4} + {^{39}}{C}_{4} \\\\ & = \left({^{36}}{C}_{5} + {^{36}}{C}_{4}\right) + {^{37}}{C}_{4} + {^{38}}{C}_{4} + {^{39}}{C}_{4} \\\\ & = \left({^{37}}{C}_{5} + {^{37}}{C}_{4}\right) + {^{38}}{C}_{4} + {^{39}}{C}_{4} \\\\ & = \left({^{38}}{C}_{5} + {^{38}}{C}_{4}\right) + {^{39}}{C}_{4} \\\\ & = {^{39}}{C}_{5} + {^{39}}{C}_{4} \\\\ & = {^{40}}{C}_{5} = RHS \\\\ & \left[\text{Note: } {^{n}}{C}_{r} + {^{n}}{C}_{\left(r - 1\right)} = {^{\left(n + 1\right)}}{C}_{r}\right] \end{aligned}$