Let m be smallest positive integer such that coefficient of x^2 in expansion of (1+x)^2+(1+x)^3+·s+

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7.Binomial Theorem

normal

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

$3$

$2$

$5$

$4$

(IIT-2016)

Solution

Coefficient of $x^2$ in expansion

$=1+{ }^3 C_2+{ }^4 C_2+.{ }^5 C_2+\ldots \ldots+{ }^{49} C_2+$

${ }^{50} C_2 \cdot m^2$

$\text { [as } .^n C_r+.{ }^n C_{r-1}=.{ }^{n+1} C_r \text { ] }$

$=\left(.{ }^3 C_5+.{ }^3 C_2\right)+.{ }^4 C_2+.{ }^5 C_2+\ldots .+.{ }^{49} C_2+$

${ }^{50} C_2 m^2$

$=\left(.{ }^4 C_3+.{ }^4 C_2\right)+\ldots . .+{ }^{50} C_2 m^2$

$=.{ }^5 C_3+.{ }^{50} C_2 m^2+.{ }^{50} C_2 m^2$

$=.{ }^{50} C_3+.{ }^{50} C_2 m^2+.{ }^{50} C_2-.{ }^{50} C_2$

$=.{ }^{51} C_3+.{ }^{50} C_2\left(m^2-1\right)$

$=(3 n+1) .{ }^{51} C_3 \text { (given) }$

$\therefore 3 n . \frac{51}{3} .{ }^{50} C_2=.{ }^{50} C_2\left(m^2-1\right)$

$\frac{m^2-1}{51}=n$

Value of $n$ is $5$ .

Standard 11

Mathematics

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hard

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normal

(JEE MAIN-2022)

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

Solution

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