A) p + 1
B) p
C) p - 1
D) p + 2
Correct Answer: B
Solution :
We have, BC = CB and \[A=B+C\Rightarrow {{A}^{p+1}}={{(B+C)}^{p+1}}\] \[{{=}^{p+1}}{{C}_{0}}{{B}_{p+1}}{{+}^{p+1}}{{C}_{1}}{{B}^{p}}C{{+}^{p+1}}{{C}_{2}}{{B}^{p-1}}{{C}^{2}}+\]?.+\[^{p+1}{{C}_{r}}{{B}^{p-1-r}}{{C}^{2}}+\]??. But\[{{C}^{2}}=O\Rightarrow {{C}^{3}}={{C}^{4}}=....={{C}^{r}}=O.\] \[\therefore \]\[{{A}^{p+1}}{{=}^{p+1}}{{C}_{0}}{{B}^{p+1}}{{+}^{p+1}}{{C}_{1}}{{B}^{p}}C\] \[={{B}^{p+1}}+(p+1){{B}^{p}}C={{B}^{p}}[B+(p+1)C]\] Thus, k = p
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