JEE Advanced 2016 · Paper 1 · Mathematics Q51

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MathematicsInteger+3Question 51 of 54

Let $m$ be the smallest positive integer such that the coefficient of $x^2$ in the expansion of $(1+x)^2+(1+x)^3+\cdots+(1+x)^{49}+(1+mx)^{50}$ is $(3n+1)\,{}^{51}C_3$ for some positive integer $n$ . Then the value of $n$ is

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Answer and worked solution

Official answer: 5

Coefficient of $x^2$ . By the hockey-stick identity,
$\sum_{r=2}^{49}\binom r2+\binom{50}2m^2=\binom{50}3+\binom{50}2m^2.$

Compare. Set this equal to $(3n+1)\binom{51}3$ and use $\binom{51}3=\binom{50}3+\binom{50}2$ :
$\binom{50}2(m^2-1)=3n\binom{51}3\implies1225(m^2-1)=62475n\implies m^2-1=51n.$

Smallest m. We need $m^2\equiv1\pmod{51}$ with $m>1$ . The first such $m$ is 16, giving $255=51\cdot5$ .

Answer: 5

Topics:Properties of binomial coefficients

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JEE Advanced 2016 · Paper 1, Mathematics question 51