A) \[^{n+m+1}{{C}_{n+1}}\]
B) \[^{n+m+2}{{C}_{n}}\]
C) \[^{n+m+3}{{C}_{n-1}}\]
D) 0
E) none of these
Correct Answer: A
Solution :
\[\sum\limits_{r=0}^{m}{^{n+r}{{C}_{n}}}=\sum\limits_{r=0}^{m}{^{n+r}{{C}_{r}}}\] \[{{=}^{n}}{{C}_{0}}{{+}^{n+1}}{{C}_{1}}{{+}^{n+2}}{{C}_{2}}+....{{+}^{n+m}}{{C}_{m}}\] \[{{=}^{n+1}}{{C}_{0}}{{+}^{n+1}}{{C}_{1}}{{+}^{n+2}}{{C}_{2}}+....{{+}^{n+m}}{{C}_{m}}\] \[[{{\because }^{n+1}}{{C}_{0}}{{=}^{n}}{{C}_{0}}]\] \[{{=}^{n+2}}{{C}_{1}}{{+}^{n+2}}{{C}_{2}}+....{{+}^{n+m}}{{C}_{m}}\] \[{{=}^{n+m}}{{C}_{m-1}}{{+}^{n+m}}{{C}_{m}}\] \[{{(}^{n}}{{C}_{r-1}}{{+}^{n}}{{C}_{r}}{{=}^{n+1}}{{C}_{r}})\] \[{{=}^{n+m+1}}{{C}_{m}}\] \[(\because {{\,}^{n}}{{C}_{r}}\,{{=}^{n}}{{C}_{n-r}})\] \[{{=}^{n+m+1}}{{C}_{n+1}}\]
You need to login to perform this action.
You will be redirected in
3 sec
