A) \[{{n}^{2}}+n-1\]
B) \[0\]
C) \[^{n+3}{{C}_{r+3}}\]
D) \[^{n}{{C}_{r-1}}{{+}^{n}}{{C}_{r}}{{+}^{n}}{{C}_{r+1}}\]
Correct Answer: B
Solution :
[b] \[\Delta =\left| \begin{matrix} ^{n}{{C}_{r-1}} & ^{n}{{C}_{r}} & (r+1){{\times }^{n+2}}{{C}_{r+1}} \\ ^{n}{{C}_{r}} & ^{n}{{C}_{r+1}} & (n+2){{\times }^{n+2}}{{C}_{r+2}} \\ ^{n}{{C}_{r+1}} & ^{n}{{C}_{r+2}} & (r+3){{\times }^{n+2}}{{C}_{r+3}} \\ \end{matrix} \right|\] Applying \[{{C}_{1}}\to {{C}_{1}}+{{C}_{2}}\] and using \[^{n}{{C}_{r}}=\frac{n}{r}{{\,}^{n-1}}{{C}_{r-1}}\]in \[{{C}_{3}},\] we get \[\Delta =\left| \begin{matrix} ^{n+1}{{C}_{r}} & ^{n}{{C}_{r}} & (n+2)\times {{\,}^{n-1}}{{C}_{r}} \\ ^{n+1}{{C}_{r+1}} & ^{n}{{C}_{r+1}} & (n+2)\times {{\,}^{n+1}}{{C}_{r+1}} \\ ^{n+1}{{C}_{r+2}} & ^{n}{{C}_{r+2}} & (n+2)\times {{\,}^{n+1}}{{C}_{r+2}} \\ \end{matrix} \right|\] \[=(n+2)\,\,\left| \begin{matrix} ^{n+1}{{C}_{r}} & ^{n}{{C}_{r}} & ^{n+1}{{C}_{r}} \\ ^{n+1}{{C}_{r+1}} & ^{n}{{C}_{r+1}} & ^{n+1}{{C}_{r+1}} \\ ^{n+1}{{C}_{r+2}} & ^{n}{{C}_{r+2}} & ^{n+1}{{C}_{r+2}} \\ \end{matrix} \right|\] \[=0\] (As \[{{C}_{1}}\] and \[{{C}_{3}}\] are identical)
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