Step-by-step Derivations

1. Identify the Expression and the Goal

We need to find the coefficient of $x^2$ in the expansion of the given expression: $S = \underbrace{{\left( {1 + x} \right)^2} + {\left( {1 + x} \right)^3} + \dots + {\left( {1 + x} \right)^{49}}}_{\text{Part 1}} + \underbrace{{{\left( {1 + mx} \right)}^{50}}}_{\text{Part 2}}$ The problem states this coefficient is equal to $(3n+1){}^{51}C_3$ for some positive integer $n$, where $m$ is the smallest positive integer for this to hold. Our goal is to find the value of $n$.

2. Find the Coefficient of $x^2$ from Part 1

Part 1 is the sum $S_1 = (1+x)^2 + (1+x)^3 + \dots + (1+x)^{49}$. The coefficient of $x^2$ in the expansion of $(1+x)^k$ is given by the binomial coefficient ${}^kC_2$. So, the coefficient of $x^2$ in $S_1$ is the sum of the individual coefficients: $C_1 = {}^2C_2 + {}^3C_2 + {}^4C_2 + \dots + {}^{49}C_2 = \sum_{k=2}^{49} {}^kC_2$ We use the hockey-stick identity, which states that $\sum_{i=r}^n \binom{i}{r} = \binom{n+1}{r+1}$. Applying this identity with $r=2$ and $n=49$: $C_1 = \sum_{k=2}^{49} {}^kC_2 = {}^{49+1}C_{2+1} = {}^{50}C_3$

Alternatively, we can sum the geometric series: $S_1 = \frac{(1+x)^2((1+x)^{48}-1)}{(1+x)-1} = \frac{(1+x)^{50}-(1+x)^2}{x}$ $S_1 = \frac{1}{x} \left[ \left(1 + {}^{50}C_1x + {}^{50}C_2x^2 + {}^{50}C_3x^3 + \dots \right) - (1+2x+x^2) \right]$ $S_1 = \frac{1}{x} \left[ (50-2)x + ({}^{50}C_2-1)x^2 + {}^{50}C_3x^3 + \dots \right]$ $S_1 = 48 + ({}^{50}C_2-1)x + {}^{50}C_3x^2 + \dots$ The coefficient of $x^2$ is indeed ${}^{50}C_3$.

3. Find the Coefficient of $x^2$ from Part 2

Part 2 is $S_2 = (1+mx)^{50}$. Using the binomial theorem, the term containing $x^2$ is: ${}^{50}C_2 (1)^{48} (mx)^2 = {}^{50}C_2 m^2 x^2$ So, the coefficient of $x^2$ from Part 2 is $C_2 = {}^{50}C_2 m^2$.

4. Formulate the Equation

The total coefficient of $x^2$ in the original expression $S$ is $C = C_1 + C_2$. $C = {}^{50}C_3 + {}^{50}C_2 m^2$ We are given that $C = (3n+1){}^{51}C_3$. So, we have the equation: ${}^{50}C_3 + {}^{50}C_2 m^2 = (3n+1){}^{51}C_3$

5. Simplify the Equation

We use Pascal's identity: ${}^{n+1}C_r = {}^nC_r + {}^nC_{r-1}$. For $n=50$ and $r=3$, we get ${}^{51}C_3 = {}^{50}C_3 + {}^{50}C_2$. Substitute this into our equation: ${}^{50}C_3 + {}^{50}C_2 m^2 = (3n+1)({}^{50}C_3 + {}^{50}C_2)$ ${}^{50}C_3 + {}^{50}C_2 m^2 = (3n+1){}^{50}C_3 + (3n+1){}^{50}C_2$ Rearranging the terms to solve for $m^2$: ${}^{50}C_2 m^2 = (3n+1-1){}^{50}C_3 + (3n+1){}^{50}C_2$ ${}^{50}C_2 m^2 = 3n \cdot {}^{50}C_3 + (3n+1){}^{50}C_2$ Now, we relate ${}^{50}C_3$ and ${}^{50}C_2$ using the identity ${}^kC_r = \frac{k-r+1}{r} {}^kC_{r-1}$. ${}^{50}C_3 = \frac{50-3+1}{3} {}^{50}C_2 = \frac{48}{3} {}^{50}C_2 = 16 \cdot {}^{50}C_2$ Substitute this back into the equation: ${}^{50}C_2 m^2 = 3n \cdot (16 \cdot {}^{50}C_2) + (3n+1){}^{50}C_2$ Since ${}^{50}C_2 \neq 0$, we can divide the entire equation by ${}^{50}C_2$: $m^2 = 48n + (3n+1)$ $m^2 = 51n + 1$

6. Find the Smallest Positive Integer $m$ and the corresponding $n$

We have the Diophantine equation $m^2 = 51n + 1$, where $m$ and $n$ are positive integers. We are looking for the smallest possible value of $m$. Rearranging, we get $m^2 - 1 = 51n$, which means $(m-1)(m+1) = 51n$. This implies that $(m-1)(m+1)$ must be a multiple of $51 = 3 \times 17$. Since $m$ must be a positive integer, $m \ge 1$. If $m=1$, $m^2-1=0$, which gives $n=0$, but $n$ must be a positive integer. So $m>1$. We test values of $m$ starting from $m=2$ to find the smallest $m$ for which $m^2-1$ is divisible by 51.

• $m=2: m^2-1=3$ (not divisible by 51)
• ...
• $m=15: m^2-1=224$ (not divisible by 51)
• $m=16: m^2-1=256-1=255$. Let's check if 255 is divisible by 51. $255 = 5 \times 51$. Yes.

For $m=16$, we have $255 = 51n$, which gives $n=5$. Since $n=5$ is a positive integer, $m=16$ is a possible value. To confirm it is the smallest, we can check systematically. We need $(m-1)(m+1)$ to be a multiple of $3 \times 17$. There are four cases for the factors $(m-1)$ and $(m+1)$:

1. $m-1$ is a multiple of 51 $\implies m-1=51k \implies m=51k+1$. Smallest $m>1$ is for $k=1$, $m=52$.
2. $m+1$ is a multiple of 51 $\implies m+1=51k \implies m=51k-1$. Smallest $m>1$ is for $k=1$, $m=50$.
3. $m-1$ is a multiple of 3 and $m+1$ is a multiple of 17. So $m \equiv 1 \pmod{3}$ and $m \equiv -1 \equiv 16 \pmod{17}$. Using Chinese Remainder Theorem or substitution: $m=17k+16$. For $k=0$, $m=16$. Check mod 3: $16 \equiv 1 \pmod{3}$. This works. So $m=16$ is a solution.
4. $m-1$ is a multiple of 17 and $m+1$ is a multiple of 3. So $m \equiv 1 \pmod{17}$ and $m \equiv -1 \equiv 2 \pmod{3}$. $m=17k+1$. $17k+1 \equiv 2 \pmod{3} \implies -k+1 \equiv 2 \implies k \equiv -1 \equiv 2 \pmod{3}$. Smallest $k \ge 0$ is $k=2$, which gives $m=17(2)+1=35$.

Comparing the possible values for $m$ (52, 50, 16, 35, ...), the smallest positive integer is $m=16$.

7. Final Answer

The problem states that $m$ is the smallest such positive integer. We found this to be $m=16$. The corresponding value of $n$ is found from the equation: $16^2 = 51n + 1$ $256 = 51n + 1$ $255 = 51n$ $n = \frac{255}{51} = 5$ The value of $n$ is 5.

Final Answer: The final answer is $\boxed{5}$.