Question

# Assertion :For $$n\ge 4$$. Let $$\displaystyle{ a }_{ n }=\sum _{ j=1 }^{ n }{ \sum _{ k=j }^{ n }{ \begin{pmatrix} n \\ k \end{pmatrix} } } \begin{pmatrix} k \\ j \end{pmatrix}$$ and $${ b }_{ n }={ a }_{ n }+{ 2 }^{ n+1 }$$;$${ b }_{ n+1 }=5{ b }_{ n }-6{ b }_{ n-1 }$$ for all $$n\ge 2$$ Reason: $${ a }_{ n }=5{ a }_{ n }-6{ a }_{ n }$$ for all $$n\ge 2$$

A
Both Assertion and Reason are correct and Reason is the correct explanation for Assertion
B
Both Assertion and Reason are correct but Reason is not the correct explanation for Assertion
C
Assertion is correct but Reason is incorrect
D
Assertion is incorrect but Reason is correct

Solution

## The correct option is A Both Assertion and Reason are correct and Reason is the correct explanation for Assertion$$\displaystyle \begin{pmatrix} n \\ k \end{pmatrix}\begin{pmatrix} k \\ j \end{pmatrix}=\frac { n! }{ k!\left( n-k \right) ! } .\frac { k! }{ j!\left( k-j \right) ! }$$$$\displaystyle=\frac { n! }{ j!\left( n-j \right) ! } \frac { \left( n-j \right) ! }{ \left( n-k \right) !\left( k-j \right) ! } =\begin{pmatrix} n \\ k \end{pmatrix}\begin{pmatrix} n\quad - & j \\ n\quad - & k \end{pmatrix}$$Thus $$\displaystyle { a }_{ n }=\sum _{ j=1 }^{ n }{ \begin{pmatrix} h \\ k \end{pmatrix} } \sum _{ j=1 }^{ n }{ \begin{pmatrix} n\quad - & j \\ n\quad - & k \end{pmatrix} }$$$$\displaystyle=\sum _{ j=1 }^{ n }{ \begin{pmatrix} n \\ j \end{pmatrix} } { 2 }^{ n-j }={ \left( 2+1 \right) }^{ n }-{ 2 }^{ n }={ 3 }^{ n }-{ 2 }^{ n }$$$$\Rightarrow { b }_{ 1 }={ 3 }^{ n }{ -2 }^{ n }+2\left( { 2 }^{ n } \right) ={ 3 }^{ n }+{ 2 }^{ n }$$Now $${ a }_{ n+1 }={ 3 }^{ n+1 }-{ 2 }^{ n+1 }=\left( 3+2 \right) \left( { 3 }^{ n }-{ 2 }^{ n } \right) -2\left( { 3 }^{ n } \right) +3\left( { 2 }^{ n } \right)$$$$=5\left( { 3 }^{ n }-{ 2 }^{ n } \right) -6\left( { 3 }^{ n-1 }-{ 2 }^{ n+1 } \right) ={ 5a }_{ n }-{ 6a }_{ n-1 }$$And $${ 5b }_{ n }-{ 6b }_{ n-1 }=5\left( { a }_{ n }+{ 2 }^{ n+1 } \right) -6\left( { a }_{ n-1 }+{ 2 }^{ n } \right)$$$$={ 5a }_{ n }-6{ b }_{ n-1 }+{ 2 }^{ n }\left( 10-6 \right)$$$$\Rightarrow { a }_{ n+1 }+{ 2 }^{ n+2 }={ b }_{ n+1 }$$Mathematics

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