Adriana´s room contains N identical light bulbs with independent switc

Our "official solution" (similar to the previous contributions) is the following:


\(?\,\,\,:\,\,\,N \ge 4\,\,{\mathop{\rm int}} \,\,{\rm{such}}\,\,{\rm{that}}\,\,\,B\left( N \right) = 26\,,\,\,\,{\rm{where}}\,\,\,B\left( N \right) = C\left( {N,2} \right) + C\left( {N,3} \right) + \ldots + C\left( {N,N} \right)\,\,\,\,\,\,\,\left( * \right)\)


\(\left( {\rm{A}} \right)\,\,\,N = 4\,\,\,\,\, \Rightarrow \,\,\,\,B\left( 4 \right)\,\, = \,\,C\left( {4,2} \right) + C\left( {4,3} \right) + C\left( {4,4} \right)\,\, = \,\,{{4 \cdot 3} \over 2} + 4 + 1\,\, = \,\,11\,\,\,\,\,\,\,\left\langle {{\rm{NO}}} \right\rangle\)

\(\left( {\rm{B}} \right)\,\,\,N = 5\,\,\,\,\, \Rightarrow \,\,\,\,B\left( 5 \right)\,\, = \,\,C\left( {5,2} \right) + C\left( {5,3} \right) + C\left( {5,4} \right) + C\left( {5,5} \right)\,\, = \,\,2 \cdot \left( {{{5 \cdot 4 \cdot 3} \over {3 \cdot 2}}} \right) + 5 + 1\,\, = \,\,20 + 5 + 1\,\,\,\,\,\,\,\left\langle {{\rm{YES}}\,{\rm{!}}} \right\rangle\)

The correct answer is therefore (B).


We follow the notations and rationale taught in the GMATH method.

Regards,
Fabio.

Bunuel
Is there any conventional approach to this question (without plugging in)?