Given that $^{n} C_{n - r} + 3^{n} C_{n - r + 1} + 3.^{n} C_{n - r + 2} +^{n} C_{n - r + 3} =^{x} C_{r}$ . Let $x = n + k$ , then find $k$ ?

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Solution

We have
$^{n} C_{n - r} + 3^{n} C_{n - r + 1} + 3.^{n} C_{n - r + 2} +^{n} C_{n - r + 3} =^{x} C_{r}$
$= (^{n} C_{n - r} +^{n} C_{n - r + 1}) + 2 (^{n} C_{n - r + 1} +^{n} C_{n - r + 2}) + (^{n} C_{n - r + 2} +^{n} C_{n - r + 3}) =^{x} C_{r}$
$=^{n + 1} C_{n - r + 1} + 2.^{n + 1} C_{n - r + 2} +^{n + 1} C_{n - r + 3} =^{x} C_{r}$
$= (^{n + 1} C_{n - r + 1} +^{n + 1} C_{n - r + 2}) + (^{n + 1} C_{n - r + 2} +^{n + 1} C_{n - r + 3}) =^{x} C_{r}$
$=^{n + 2} C_{n - r + 2} +^{n + 2} C_{n - r + 3} =^{x} C_{r}$
$=^{n + 3} C_{n - r + 3} =^{x} C_{r}$
$=^{n + 3} C_{r} =^{x} C_{r}$
Hence, $x = n + 3$
$∴ k = 3$

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