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Binomial Coefficients

Let m be the ...

Question

Let $m$ be the smallest positive integer such that the coefficient of $x^{2}$ in the expansion of $(1 + x)^{2} + (1 + x)^{3} + . . . + (1 + x)^{49} + (1 + m x)^{50} is (3 n + 1)^{51} C_{3}$ for some positive integer $n$ . Then the
value of $n$ is

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Solution

$(1 + x)^{2} + (1 + x)^{3} + . . . + (1 + x)^{49} + (1 + m x)^{50}$

$Coefficient of x^{2} in the above expression =$
$^{2} C_{2} +^{3} C_{2} +^{4} C_{2} + \dots +^{49} C_{2} +^{50} C_{2} m^{2} = (3 n + 1)^{51} C_{3}$
$^{3} C_{3} +^{3} C_{2} +^{4} C_{2} + \dots +^{49} C_{2} +^{50} C_{2} m^{2} = (3 n + 1)^{51} C_{3}$
As $^{n} C_{r} +^{n} C_{r - 1} =^{n + 1} C_{r}$
$∴^{50} C_{3} +^{50} C_{2} m^{2} = (3 n + 1)^{51} C_{3} \Rightarrow \frac{^{50} C_{3}}{^{51} C_{3}} + \frac{^{50} C_{2}}{^{51} C_{3}} m^{2} = 3 n + 1 \Rightarrow \frac{16}{17} + \frac{1}{17} m^{2} = 3 n + 1 \Rightarrow m^{2} = 51 n + 1 \Rightarrow n = \frac{m^{2} - 1}{51} = \frac{(m + 1) (m - 1)}{51}$
So, the numerator should be divisible by $17$ and $3$ for $n$ to be a integer.
Smallest value of $m$ for which $n$ is an integer is, $m = 16$ for which $n = 5$

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