A) 1
B) -1
C) 0
D) none of these
Correct Answer: A
Solution :
[a] \[\left| \begin{matrix} 1 & 1 & 1 \\ ^{m}{{C}_{1}} & ^{m+1}{{C}_{1}} & ^{m+2}{{C}_{1}} \\ ^{m}{{C}_{2}} & ^{m+1}{{C}_{2}} & ^{m+2}{{C}_{2}} \\ \end{matrix} \right|\]=\[\left| \begin{matrix} 1 & 1 & 1 \\ ^{m}{{C}_{1}} & ^{m+1}{{C}_{1}} & ^{m+1}{{C}_{0}}{{+}^{m+2}}{{C}_{1}} \\ ^{m}{{C}_{2}} & ^{m+1}{{C}_{2}} & ^{m+1}{{C}_{1}}{{+}^{m+1}}{{C}_{2}} \\ \end{matrix} \right|\]\[\left| \begin{matrix} 1 & 1 & 0 \\ ^{m}{{C}_{1}} & ^{m+1}{{C}_{1}} & ^{m+1}{{C}_{0}} \\ ^{m}{{C}_{2}} & ^{m+1}{{C}_{2}} & ^{m+1}{{C}_{1}} \\ \end{matrix} \right|\] [Applying\[{{C}_{3}}\to {{C}_{3}}-{{C}_{2}}\]] \[=\left| \begin{matrix} 1 & 1 & 0 \\ ^{m}{{C}_{1}} & ^{m}{{C}_{0}}{{+}^{m}}{{C}_{1}} & ^{m+1}{{C}_{0}} \\ ^{m}{{C}_{2}} & ^{m}{{C}_{1}}{{+}^{m}}{{C}_{2}} & ^{m+1}{{C}_{1}} \\ \end{matrix} \right|\] \[=\left| \begin{matrix} 1 & 0 & 0 \\ ^{m}{{C}_{1}} & ^{m}{{C}_{0}} & ^{m+1}{{C}_{0}} \\ ^{m}{{C}_{2}} & ^{m}{{C}_{1}} & ^{m+1}{{C}_{1}} \\ \end{matrix} \right|\] [Applying\[{{C}_{2}}\to {{C}_{2}}-{{C}_{1}}\]] \[={}^{m}{{C}_{0}}^{m+1}{{C}_{1}}-{}^{m+1}{{C}_{0}}^{m}{{C}_{1}}=m+1-m=1\]
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