Let m be the smallest positive integer such that the coefficient of ${{\text{x}}^{\text{2}}}$ in the expansion of ${\left( {{\text{1 + x}}} \right)^{\text{2}}}{\text{ + }}{\left( {{\text{1 + x}}} \right)^{\text{3}}}{\text{ + }}{\left( {{\text{1 + x}}} \right)^{\text{4}}}{\text{ + }}...........{\text{ + }}{\left( {{\text{1 + x}}} \right)^{{\text{49}}}}{\text{ + }}{\left( {{\text{1 + x}}} \right)^{{\text{50}}}}$ is (3n+1)${}^{{\text{51}}}{{\text{C}}_{\text{3}}}$ for some positive integer n. Then the value of n is-

Hint-To solve this question, we need to know the basics of Binomial Theorem. i.e.(x+y)n = nΣr=0 nCr xn – r · yr. where,${}^{\text{n}}{{\text{C}}_{\text{r}}}{\text{ = }}\dfrac{{{\text{n}}!}}{{{\text{[r}}!{\text{(n - r)}}!{\text{]}}}}$

Complete step-by-step answer:
And by using the above statement we will get the value of ${{\text{x}}^{\text{2}}}$ in the given expansion.
Now, from above equation, we observe that,
${\left( {{\text{1 + x}}} \right)^{\text{2}}}$$ \to $coefficient of ${{\text{x}}^{\text{2}}}$= ${}^{\text{2}}{{\text{C}}_{\text{2}}}$
${\left( {{\text{1 + x}}} \right)^3}$$ \to $coefficient of ${{\text{x}}^{\text{2}}}$= ${}^3{{\text{C}}_{\text{2}}}$
${\left( {{\text{1 + x}}} \right)^4}$$ \to $coefficient of ${{\text{x}}^{\text{2}}}$= ${}^4{{\text{C}}_{\text{2}}}$
Similarly, we say,
${\left( {{\text{1 + x}}} \right)^{\text{n}}}$$ \to $coefficient of ${{\text{x}}^{\text{2}}}$= ${}^n{{\text{C}}_{\text{2}}}$
According to question:
${\left( {{\text{1 + x}}} \right)^{\text{2}}}{\text{ + }}{\left( {{\text{1 + x}}} \right)^{\text{3}}}{\text{ + }}{\left( {{\text{1 + x}}} \right)^{\text{4}}}{\text{ + }}...........{\text{ + }}{\left( {{\text{1 + x}}} \right)^{{\text{49}}}}{\text{ + }}{\left( {{\text{1 + x}}} \right)^{{\text{50}}}}$$ \to $coefficient of ${{\text{x}}^{\text{2}}}$= (3n+1)${}^{{\text{51}}}{{\text{C}}_{\text{3}}}$
coefficient of ${{\text{x}}^{\text{2}}}$ in given expression-
${}^{\text{2}}{{\text{C}}_{\text{2}}}$+${}^3{{\text{C}}_{\text{2}}}$+${}^4{{\text{C}}_{\text{2}}}$+………….+${}^{49}{{\text{C}}_{\text{2}}}$+${}^{50}{{\text{C}}_{\text{2}}}$${{\text{m}}^{\text{2}}}$
As we know,
${}^{\text{n}}{{\text{C}}_{\text{r}}}{\text{ + }}{}^{\text{n}}{{\text{C}}_{{\text{r - 1}}}}{\text{ = }}{}^{{\text{n + 1}}}{{\text{C}}_{\text{r}}}$
${}^{\text{3}}{{\text{C}}_3}$+${}^{\text{3}}{{\text{C}}_{\text{2}}}$+${}^4{{\text{C}}_{\text{2}}}$+…………+${}^{49}{{\text{C}}_{\text{2}}}$+${}^{50}{{\text{C}}_{\text{2}}}$${{\text{m}}^{\text{2}}}$
${}^4{{\text{C}}_3}$+${}^4{{\text{C}}_{\text{2}}}$+${}^5{{\text{C}}_{\text{2}}}$+…………+${}^{49}{{\text{C}}_{\text{2}}}$+${}^{50}{{\text{C}}_{\text{2}}}$${{\text{m}}^{\text{2}}}$
${}^5{{\text{C}}_3}$+${}^5{{\text{C}}_{\text{2}}}$+………
${}^6{{\text{C}}_3}$+${}^6{{\text{C}}_{\text{2}}}$+${}^7{{\text{C}}_2}$……
Similarly, by multiple simplification, we get,
${}^{49}{{\text{C}}_3}$+${}^{49}{{\text{C}}_{\text{2}}}$+${}^{50}{{\text{C}}_{\text{2}}}$${{\text{m}}^{\text{2}}}$
${}^{50}{{\text{C}}_3}$+ (${}^{50}{{\text{C}}_{\text{2}}}$${{\text{m}}^{\text{2}}}$-${}^{50}{{\text{C}}_{\text{2}}}$)+${}^{50}{{\text{C}}_{\text{2}}}$
${}^{51}{{\text{C}}_3}$+${}^{50}{{\text{C}}_{\text{2}}}$(${{\text{m}}^{\text{2}}}$-1)
Now, according to question-
${}^{51}{{\text{C}}_3}$+${}^{50}{{\text{C}}_{\text{2}}}$(${{\text{m}}^{\text{2}}}$-1) = (3n+1)${}^{{\text{51}}}{{\text{C}}_{\text{3}}}$
${}^{51}{{\text{C}}_3}$+${}^{50}{{\text{C}}_{\text{2}}}$(${{\text{m}}^{\text{2}}}$-1) = 3n${}^{51}{{\text{C}}_3}$+${}^{51}{{\text{C}}_3}$
${}^{50}{{\text{C}}_{\text{2}}}$(${{\text{m}}^{\text{2}}}$-1) = 3n$\dfrac{{51!}}{{{\text{[3}}!{\text{(51 - 3)}}!{\text{]}}}}$
3n$ \times \dfrac{{51}}{3}$$\dfrac{{50!}}{{{\text{[2}}!{\text{(51 - 2)}}!{\text{]}}}}$= ${}^{50}{{\text{C}}_{\text{2}}}$(${{\text{m}}^{\text{2}}}$-1)

3n$ \times \dfrac{{51}}{3}$$ \times $${}^{50}{{\text{C}}_2}$= ${}^{50}{{\text{C}}_{\text{2}}}$(${{\text{m}}^{\text{2}}}$-1)
51n+1=${{\text{m}}^{\text{2}}}$
Since, m be the smallest positive integer for some positive integer n.
And select n in such a way that m is a perfect square.
So, if we take n=5
We get ${{\text{m}}^{\text{2}}}$= 256, which is a perfect square.
Thus, the value of n will be 5 in this question.

Note- The total number of terms in the expansion of (x+y)n are (n+1). The sum of exponents of x and y is always n. and The binomial coefficients which are equidistant from the beginning and from the ending are equal i.e. nC0 = nCn, nC1 = nCn-1 , nC2 = nCn-2 ,….. etc.