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

JEE Advanced · 2016 · 22 May 2016 · Paper 1 · Binomial Theorem

Let

m

be the smallest positive integer such that the coefficient of

x^{2}

in the expansion of

(1 + x)^{2} + (1 + x)^{3} + \dots + (1 + x)^{49} + (1 + m x)^{50}

(3 n + 1)^{51} C_{3}

for some positive integer

n

. Then the value of

n

Answer: 5

The coefficient of

x^{2}

(1 + x)^{2} + (1 + x)^{3} + \dots + (1 + x)^{49}

is:

k = 2 \sum 49 (2 k) = (3 50) - (3 2) = (3 50)

The coefficient of

x^{2}

(1 + m x)^{50}

(2 50) m^{2}

.

Total coefficient of

x^{2}

(3 50) + (2 50) m^{2} = (3 n + 1) (3 51)

Using

(3 51) = \frac{51 \cdot 50 \cdot 49}{6}

and

(3 50) = \frac{50 \cdot 49 \cdot 48}{6}

\frac{50 \cdot 49 \cdot 48}{6} + \frac{50 \cdot 49}{2} m^{2} = (3 n + 1) \frac{51 \cdot 50 \cdot 49}{6}

Simplifying:

48 + 3 m^{2} = 51 (3 n + 1)

3 m^{2} = 153 n + 51 - 48 = 153 n + 3

m^{2} = 51 n + 1

Smallest positive integer

m

: trying values,

n = 5

gives

m^{2} = 256

, so

m = 16

.

Answer: 5.

JEE Advanced · 2016 · 22 May 2016 · Paper 1 · Integer answer