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Multiplication of Matrices

Let Ck = ...

Question

Let $C_{k} =$ $^{n} C_{k}$ for $0 \leq k \leq n$ and
$A_{k} = [\begin{matrix} C_{k - 1}^{2} & 0 0 & C_{k}^{2} \end{matrix}]$ for $k \geq 1$ , and $A_{1} + A_{2} + . . . + A_{n} = [\begin{matrix} k_{1} & 0 0 & k_{2} \end{matrix}]$ , then

A

k_{1} = k_{2}

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B

k_{1} + k_{2} =

^{2 n} C_{2 n} + 1

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C

k_{1} =

^{2 n} C_{n} - 1

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D

k_{2} =

^{2 n} C_{n + 1}

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Solution

The correct options are
A $k_{1} = k_{2}$
C $k_{1} =$ $^{2 n} C_{n} - 1$
$C_{k} =$ $^{n} C_{k}$ for $0 \leq k \leq n$ and
$A_{k} = [\begin{matrix} C_{k - 1}^{2} & 0 0 & C_{k}^{2} \end{matrix}]$ for $k \geq 1$
$A_{1} + A_{2} + . . . + A_{n} = [\begin{matrix} k_{1} & 0 0 & k_{2} \end{matrix}]$
but we know that $n \sum k = 0 C_{k}^{2} =$ $^{2 n} C_{n}$
$\Rightarrow A_{1} + A_{2} + . . . + A_{n} = [\begin{matrix} ^{2 n} C_{n} - 1 & 0 0 & ^{2 n} C_{n} - 1 \end{matrix}]$
$\Rightarrow k_{1} = k_{2}$ and $k_{1} =^{2 n} C_{n} - 1$
Hence, options $A$ and $C$ .

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