A) 5
B) 7
C) 6
D) 4
Correct Answer: B
Solution :
The number of triangles can be formed using n non-collinear points \[{{=}^{n}}{{C}_{3}}\]. Since, \[{{T}_{n}}={{\,}^{n}}\,{{C}_{3}}\] Given, \[{{T}_{n+1}}-T{{}_{n}}=21\] \[\Rightarrow \] \[^{n+1}{{C}_{3}}{{-}^{n}}{{C}_{3}}=21\] \[{{\Rightarrow }^{n}}{{C}_{3}}{{+}^{n}}{{C}_{3}}{{-}^{n}}{{C}_{3}}=21\,({{\because }^{\,\,n}}{{C}_{2}}{{+}^{n}}{{C}_{3}}{{=}^{n+1}}{{C}_{3}})\] \[\Rightarrow \] \[^{n}{{C}_{2}}=21\] \[\Rightarrow \] \[\frac{n(n-1)}{2}=21\Rightarrow {{n}^{2}}-n-42=0\] \[\Rightarrow \] \[(n-7)\,(n+6)=0\] \[\Rightarrow \] \[n=7\] \[(\because n\ne -6)\]
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