A) \[\frac{{{C}_{p}}}{{{C}_{v}}}\]
B) \[{{C}_{v}}=\frac{{{n}_{1}}{{C}_{{{v}_{1}}}}+{{n}_{2}}{{C}_{{{v}_{2}}}}}{{{n}_{1}}+{{n}_{2}}}\]
C) 2n + 1
D) None of these
Correct Answer: B
Solution :
We have,\[{{A}^{2}}=A\] \[\therefore \]\[{{(I+A)}^{2}}=(I+A)(I+A)=I+2A+{{A}^{2}}+I+3A\]and\[{{(I+A)}^{2}}={{(I+A)}^{2}}(I+A)\] \[=(I+3A)(I+A)\]\[[\because {{(I+A)}^{2}}=I+3A]\] \[=I+4A+3{{A}^{2}}\] \[=I+7A\] \[[\because {{A}^{2}}=A]\] Thus, we have \[{{(I+A)}^{2}}=I+3A\]and\[{{(I+A)}^{3}}=I+7A\] \[\Rightarrow \]\[{{(I+A)}^{2}}=I+({{2}^{2}}-1)A\] and\[{{(I+A)}^{3}}=I+({{2}^{3}}-1)A\] Hence,\[{{(I+A)}^{n}}=I+({{2}^{n}}-1)A\] \[\therefore \]\[{{(I+A)}^{n}}+I+\lambda A\]\[\Rightarrow \]\[\lambda ={{2}^{n}}-1\]
You need to login to perform this action.
You will be redirected in
3 sec
