A) \[\sum\limits_{k=0}^{n}{{{[C\,\,(n,\,\,k)]}^{2}}}\]
B) \[\sum\limits_{k=0}^{n}{{{[C\,\,(n,\,\,k+2)]}^{2}}}\]
C) \[\sum\limits_{k=0}^{n}{{{[C\,\,(n,\,\,k+3)]}^{2}}}\]
D) \[\sum\limits_{k=0}^{n3}{{{[C\,\,(n,\,\,k)]}^{2}}}\]
Correct Answer: D
Solution :
Expression, \[{{[(1+p)(1+q)\,(p+q)]}^{n}}\] \[={{(1+p)}^{n}}\,{{(1+q)}^{n}}{{(p+q)}^{n}}\] \[={{(}^{0}}{{C}_{0}}{{+}^{n}}{{C}_{1}}p{{+}^{n}}{{C}_{2}}{{p}^{2}}+....{{+}^{n}}{{C}_{n}}{{p}^{n}})\] \[.{{(}^{n}}{{C}_{0}}{{+}^{n}}{{C}_{1}}q{{+}^{n}}{{C}_{2}}{{q}^{2}}+....{{+}^{n}}{{C}_{n}}{{q}^{n}})\] \[.{{(}^{n}}{{C}_{0}}{{p}^{n}}{{+}^{n}}{{C}_{1}}{{p}^{n-1}}q+....{{+}^{n}}{{C}_{n}}{{q}^{n}})\] Now, coefficient of \[{{p}^{n}}\,\,{{q}^{n}}\] in the above expansion \[{{=}^{n}}{{C}_{0}}\,{{\,}^{n}}{{C}_{n}}\,{{\,}^{n}}{{C}_{0}}{{+}^{n}}{{C}_{1}}{{\,}^{n}}{{C}_{n-1}}{{\,}^{n}}{{C}_{1}}{{+}^{n}}{{C}_{2}}{{\,}^{n}}{{C}_{n-2}}{{\,}^{n}}{{C}_{2}}\] \[+....{{+}^{n}}{{C}_{n}}\,{{\,}^{n}}{{C}_{0}}\,\,{{\,}^{n}}{{C}_{n}}\,\] \[{{=}^{n}}{{C}_{0}}{{\,}^{n}}{{C}_{0}}{{\,}^{n}}{{C}_{0}}{{+}^{n}}{{C}_{1}}{{\,}^{n}}{{C}_{1}}{{\,}^{n}}{{C}_{1}}{{+}^{n}}{{C}_{2}}{{\,}^{n}}{{C}_{2}}{{\,}^{n}}{{C}_{2}}\] \[+....{{+}^{n}}{{C}_{n}}{{\,}^{n}}{{C}_{n}}{{\,}^{n}}{{C}_{n}}\] \[={{{{(}^{n}}{{C}_{0}})}^{3}}+{{{{(}^{n}}{{C}_{1}})}^{3}}+{{{{(}^{n}}{{C}_{2}})}^{3}}+...+{{{{(}^{n}}{{C}_{n}})}^{3}}\] \[=\sum\limits_{k=0}^{n}{{{[C\,(n,\,\,k)]}^{3}}}\]You need to login to perform this action.
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