A) \[^{n+m+1}{{C}_{n+1}}\]
B) \[^{n+m+2}{{C}_{n}}\]
C) \[^{n+m+3}{{C}_{n-1}}\]
D) None of these
Correct Answer: A
Solution :
Since \[^{n}{{C}_{r}}{{=}^{n}}{{C}_{n-r}}\] and \[^{n}{{C}_{r-1}}{{+}^{n}}{{C}_{r}}{{=}^{n+1}}{{C}_{r}}\] we have \[\sum\limits_{r=0}^{m}{^{n+r}{{C}_{n}}}=\sum\limits_{r=0}^{m}{^{n+r}{{C}_{r}}}{{=}^{n}}{{C}_{0}}{{+}^{n+1}}{{C}_{1}}{{+}^{n+2}}{{C}_{2}}+......{{+}^{n+m}}{{C}_{m}}\] \[=[1+(n+1)]{{+}^{n+2}}{{C}_{2}}{{+}^{n+3}}{{C}_{3}}+........{{+}^{n+m}}{{C}_{m}}\] \[{{=}^{n+m+1}}{{C}_{n+1}}\], \[[\because {{\ }^{n}}{{C}_{r}}{{=}^{n}}{{C}_{n-r}}]\].
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