A) \[n\cdot {{2}^{n-1}}-1\]
B) \[n\cdot {{2}^{n-1}}+(n+1)!\]
C) \[n\cdot {{2}^{n-1}}+(n+1)!\,\,-1\]
D) \[{{n}^{2}}+n+5\]
Correct Answer: C
Solution :
\[{{t}_{r}}=r\,({}^{n}{{C}_{r}}+{}^{n}{{P}_{r}})\] \[=r\,.\,{}^{n}{{C}_{r}}+r\,.\,{}^{n}{{P}_{r}}\] \[=n\,.\,{}^{n-1}{{C}_{r-1}}+r\,.\,\,(r!)\] \[=n\,.\,{}^{n-1}{{C}_{r-1}}+\,\,(r+1)!\,-r!\] \[\therefore \] \[\text{sum}=\,n\,\{{}^{n-1}{{C}_{0}}+{}^{n-1}{{C}_{1}}+.....+{}^{n-1}{{C}_{n-1}}\}\] \[+\,\{(2!\,-1!)+(3!\,-2!)+....((n+1)!-n!)\] \[=n\,.\,\,{{2}^{n-1}}+(n+1)!\,-1!\]You need to login to perform this action.
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