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Question

Let $$C_{k}=^{n}C_{k}$$ for $$0\le k \le n$$ and $${ A }_{ k }=\begin{bmatrix} { C }_{ k-1 }^{ 2 } & 0 \\ 0 & { C }_{ k }^{ 2 } \end{bmatrix}$$ for $$k\ge 1$$ and $${ A }_{ 1 }+{ A }_{ 2 }+....{ A }_{ n }=\begin{bmatrix} { k }_{ 1 } & 0 \\ 0 & { k }_{ 2 } \end{bmatrix}$$ then


A
k1=k2
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B
k1+k2=2nC2n+1
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C
k1=2nCn1
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D
k2=2nCn1
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Solution

The correct option is A $$k_{1}=k_{2}$$
$$\begin{array}{l} We\, have \\ { A_{ k } }=\left| { \begin{array} { *{ 20 }{ c } }{ { c^{ 2 } }_{ k-1 } } & 0 \\ 0 & { c_{ 2 }^{ 2 } } \end{array} } \right|  \\ { A_{ n } }=\left[ { \begin{array} { *{ 20 }{ c } }{ c_{ 0 }^{ 2 } } & 0 \\ 0 & { c_{ 1 }^{ 2 } } \end{array} } \right] +\left[ { \begin{array} { *{ 20 }{ c } }{ c_{ 1 }^{ 2 } } & 0 \\ 0 & { c_{ 2 }^{ 2 } } \end{array} } \right] +.....+\left[ { \begin{array} { *{ 20 }{ c } }{ c_{ n-1 }^{ 2 } } & 0 \\ 0 & { c_{ n }^{ 2 } } \end{array} } \right]  \\ =\left[ { \begin{array} { *{ 20 }{ c } }{ c_{ 0 }^{ 2 }+c_{ 1 }^{ 2 }+...+c_{ n-1 }^{ 2 }{ 1pt } } & 0 \\ 0 & { c_{ 1 }^{ 2 }+c_{ 2 }^{ 2 }+...+c_{ n }^{ 2 }{ 1pt } } \end{array} } \right]  \\ =\left[ { \begin{array} { *{ 20 }{ c } }{ ^{ 2n }{ c_{ n } }-1 } & 0 \\ 0 & { ^{ 2n }{ c_{ n } }-1 } \end{array} } \right]  \\  \\ { \left( { 1+x } \right) ^{ n } }{ =^{ n } }{ c_{ 0 } }{ +^{ n } }{ c_{ 1 } }x{ +^{ n } }{ c_{ 2 } }{ x^{ 2 } }+......{ +^{ n } }{ c_{ n } }{ x^{ n } } \\ { \left( { x+1 } \right) ^{ n } }{ =^{ n } }{ c_{ 0 } }{ x^{ n } }{ +^{ n } }{ c_{ 1 } }{ x^{ n-1 } }{ +^{ n } }{ c_{ 2 } }{ x^{ x-2 } }+.......{ +^{ n } }{ c_{ n } } \\  \\ { \left( { 1+x } \right) ^{ 2n } }=\left[ { ^{ n }{ c_{ 0 } }{ +^{ n } }{ c_{ 1 } }x+....{ +^{ n } }{ c_{ n } }{ x^{ n } } } \right] \left[ { ^{ n }{ c_{ 0 } }{ x^{ n } }{ +^{ n } }{ c_{ 1 } }{ x^{ n-1 } }+....{ +^{ n } }{ c_{ n } } } \right]  \\ coefficient\, of\, { x^{ n } }:{ \, ^{ 2n } }{ c_{ n } }{ =^{ n } }c_{ 0 }^{ 2 }{ +^{ n } }c_{ 0 }^{ 2 }+....{ +^{ n } }c_{ n }^{ 2 } \\ { K_{ 1 } }={ K_{ 2 } }{ =^{ 2n } }{ c_{ n } }-1 \\ Hence,\, option\, A\, is\, the\, correct\, answer. \end{array}$$

Mathematics

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