A) \[I+P\]
B) \[I+nP\]
C) \[I+{{2}^{n}}P\]
D) \[I+({{2}^{n}}-1)P\]
Correct Answer: D
Solution :
We have, \[{{P}^{2}}=P\] \[\Rightarrow \,\,\,\,{{P}^{-1}}{{P}^{2}}={{P}^{-1}}P\] \[\Rightarrow \,\,\,\,\,P=I\] \[\Rightarrow \,\,{{\left( I+P \right)}^{n}}={{\left( P+P \right)}^{n}}\] |
\[={{\left( 2P \right)}^{n}}={{2}^{n}}{{P}^{n}}={{2}^{n}}P\] |
\[=P+{{2}^{n}}\,\,\,P-P\] |
\[=I+\left( {{2}^{n}}-1 \right)p\] |
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