A) \[{{n}^{2}}-2np+4{{p}^{2}}=0\]
B) \[{{n}^{2}}-n\,(4p+1)+4{{p}^{2}}-2=0\]
C) \[{{n}^{2}}-n\,(4p+1)+4{{p}^{2}}=0\]
D) None of these
Correct Answer: B
Solution :
Coefficient of \[{{p}^{th}},{{(p+1)}^{th}}\]and \[{{(p+2)}^{th}}\] terms in expansion of \[{{(1+x)}^{n}}\] are \[^{n}{{C}_{p-1}}{{,}^{n}}{{C}_{p}}{{,}^{n}}{{C}_{p+1}}\]. Then \[{{2}^{n}}{{C}_{p}}={{\,}^{n}}{{C}_{p-1}}+{{\,}^{n}}{{C}_{p+1}}\] \[\Rightarrow {{n}^{2}}-n(4p+1)+4{{p}^{2}}-2=0\] Trick: Let p = 1, hence \[^{n}{{C}_{0}},{{\,}^{n}}{{C}_{1}}\]and \[^{n}{{C}_{2}}\]are in A.P. \[\Rightarrow \,\,\,{{2.}^{n}}{{C}_{1}}={{\,}^{n}}{{C}_{0}}{{+}^{n}}{{C}_{2}}\,\,\,\,\Rightarrow 2n=1+\frac{n\,(n-1)}{2}\] \[\Rightarrow \,\,4n=2+{{n}^{2}}-n\,\,\,\,\Rightarrow {{n}^{2}}-5n+2=0\] which is given by B.
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