A) 1
B) ½
C) 2
D) None of these
Correct Answer: A
Solution :
[a] We have, BC = CB, and \[{{A}^{N+1}}={{(B+C)}^{N+1}}\] \[{{=}^{N+1}}{{C}_{0}}{{B}^{N+1}}{{+}^{N+1}}{{C}_{1}}\,{{B}^{N}}C{{+}^{N+1}}{{C}_{2}}{{B}^{N-1}}{{C}^{2}}+\] \[....+{{\,}^{N+1}}{{C}_{r}}{{B}^{N+1-r}}{{C}_{r}}+.....\] But given that \[{{C}^{2}}=0\Rightarrow {{C}^{3}}={{C}^{4}}=...={{C}^{r}}=0\] Hence, \[{{A}^{N+1}}={{\,}^{N+1}}{{C}_{N}}{{B}^{N+1}}+{{\,}^{N+1}}{{C}_{1}}{{B}^{N}}C\] \[={{B}^{N+1}}+(N+1){{B}^{N}}C={{B}^{N}}[B+(N+1)C]\] Thus \[K=N\]
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