Key Concepts and Formulas

• Variance ($\sigma^2$): A statistical measure quantifying the extent to which data points diverge from the mean. For a set of $N$ observations $x_1, x_2, \dots, x_N$, the most efficient formula is: $\sigma^2 = E[X^2] - (E[X])^2$ where $E[X] = \frac{\sum x_i}{N}$ is the mean, and $E[X^2] = \frac{\sum x_i^2}{N}$ is the mean of the squares.
• Summation Formulas:
  • Sum of the first $N$ natural numbers: $\sum_{i=1}^N i = \frac{N(N+1)}{2}$
  • Sum of the squares of the first $N$ natural numbers: $\sum_{i=1}^N i^2 = \frac{N(N+1)(2N+1)}{6}$
• Property of Variance for Linear Transformations: If $Y = aX + b$, then $\text{Var}(Y) = a^2 \text{Var}(X)$. This means that adding a constant ($b$) does not affect variance, but multiplying by a constant ($a$) scales the variance by $a^2$.

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Step-by-Step Solution

Part 1: Variance of the First $n$ Natural Numbers

We are given that the variance of the first $n$ natural numbers ($1, 2, 3, \dots, n$) is $10$. Our goal is to determine the value of $n$.

• Step 1.1: Calculate the Mean ($E[X]$) for the first $n$ natural numbers. Why this step? The variance formula $\sigma^2 = E[X^2] - (E[X])^2$ requires the mean, $E[X]$. The series is $x_i = \{1, 2, \dots, n\}$, with $N=n$ observations. Sum of the first $n$ natural numbers: $\sum_{i=1}^n x_i = \sum_{i=1}^n i = \frac{n(n+1)}{2}$. Mean: $E[X] = \frac{\sum x_i}{N} = \frac{n(n+1)/2}{n} = \frac{n+1}{2}$

• Step 1.2: Calculate the Mean of Squares ($E[X^2]$) for the first $n$ natural numbers. Why this step? The variance formula also requires the mean of squares, $E[X^2]$. Sum of the squares of the first $n$ natural numbers: $\sum_{i=1}^n x_i^2 = \sum_{i=1}^n i^2 = \frac{n(n+1)(2n+1)}{6}$. Mean of squares: $E[X^2] = \frac{\sum x_i^2}{N} = \frac{n(n+1)(2n+1)/6}{n} = \frac{(n+1)(2n+1)}{6}$

• Step 1.3: Apply the Variance Formula and Solve for $n$. Why this step? We substitute the calculated $E[X]$ and $E[X^2]$ into the variance formula and equate it to the given variance of $10$ to find $n$. $\sigma^2 = E[X^2] - (E[X])^2$ $10 = \frac{(n+1)(2n+1)}{6} - \left(\frac{n+1}{2}\right)^2$ Factor out the common term $(n+1)$ to simplify the algebra: $10 = (n+1) \left[ \frac{2n+1}{6} - \frac{n+1}{4} \right]$ Find a common denominator (12) for the terms inside the bracket: $10 = (n+1) \left[ \frac{2(2n+1) - 3(n+1)}{12} \right]$ $10 = (n+1) \left[ \frac{4n+2 - 3n-3}{12} \right]$ $10 = (n+1) \left[ \frac{n-1}{12} \right]$ Recognize the difference of squares: $(n+1)(n-1) = n^2-1$. $10 = \frac{n^2-1}{12}$ $120 = n^2-1$ $n^2 = 121$ Taking the square root: $n = \pm 11$. Since $n$ represents the number of natural numbers, it must be a positive integer. Therefore, $n = 11$.

Part 2: Variance of the First $m$ Even Natural Numbers

We are given that the variance of the first $m$ even natural numbers ($2, 4, 6, \dots, 2m$) is $16$. Our goal is to determine the value of $m$. We will demonstrate two methods.

Method 1: Direct Calculation (Similar to Part 1)

• Step 2.1: Calculate the Mean ($E[X]$) for the first $m$ even natural numbers. The series is $x_i = \{2, 4, \dots, 2m\}$, with $N=m$ observations. Sum of the first $m$ even natural numbers: $\sum_{i=1}^m x_i = \sum_{i=1}^m (2i) = 2 \sum_{i=1}^m i = 2 \times \frac{m(m+1)}{2} = m(m+1)$. Mean: $E[X] = \frac{m(m+1)}{m} = m+1$

• Step 2.2: Calculate the Mean of Squares ($E[X^2]$) for the first $m$ even natural numbers. Sum of the squares: $\sum_{i=1}^m x_i^2 = \sum_{i=1}^m (2i)^2 = \sum_{i=1}^m 4i^2 = 4 \sum_{i=1}^m i^2$. Using the sum of squares formula: $4 \times \frac{m(m+1)(2m+1)}{6} = \frac{2m(m+1)(2m+1)}{3}$. Mean of squares: $E[X^2] = \frac{2m(m+1)(2m+1)/3}{m} = \frac{2(m+1)(2m+1)}{3}$

• Step 2.3: Apply the Variance Formula and Solve for $m$. Substitute $E[X^2]$ and $E[X]$ into the variance formula, equating it to $16$. $16 = \frac{2(m+1)(2m+1)}{3} - (m+1)^2$ Factor out $(m+1)$: $16 = (m+1) \left[ \frac{2(2m+1)}{3} - (m+1) \right]$ Find a common denominator (3) for the terms inside the bracket: $16 = (m+1) \left[ \frac{4m+2 - 3(m+1)}{3} \right]$ $16 = (m+1) \left[ \frac{4m+2 - 3m - 3}{3} \right]$ $16 = (m+1) \left[ \frac{m-1}{3} \right]$ Using the difference of squares: $16 = \frac{m^2-1}{3}$ $48 = m^2-1$ $m^2 = 49$ Taking the square root: $m = \pm 7$. Since $m$ represents the number of even natural numbers, it must be a positive integer. Therefore, $m = 7$.

Method 2: Using Properties of Variance (More Efficient)

Why use this method? This method leverages a powerful property of variance to simplify calculations, especially when data is a linear transformation of another dataset whose variance is known.

• Step 2.4: Express the series as a linear transformation. The series of first $m$ even natural numbers is $\{2, 4, 6, \dots, 2m\}$. This can be written as $Y_i = 2 \times X_i$, where $X_i = \{1, 2, 3, \dots, m\}$ are the first $m$ natural numbers. Here, the scaling factor $a=2$ and the constant shift $b=0$.

• Step 2.5: Apply the variance property. From Part 1, we know the variance of the first $k$ natural numbers is $\frac{k^2-1}{12}$. Applying the property $\text{Var}(aX+b) = a^2 \text{Var}(X)$: $\text{Var}(\text{first } m \text{ even natural numbers}) = \text{Var}(2 \times \text{first } m \text{ natural numbers})$ $= 2^2 \times \text{Var}(\text{first } m \text{ natural numbers})$ $= 4 \times \left( \frac{m^2-1}{12} \right)$ $= \frac{m^2-1}{3}$

• Step 2.6: Solve for $m$. We are given that this variance is $16$. $16 = \frac{m^2-1}{3}$ $48 = m^2-1$ $m^2 = 49$ Taking the square root: $m = \pm 7$. Since $m$ must be positive, $m = 7$.

Part 3: Final Calculation of $m+n$

• Step 3.1: Sum the values of $m$ and $n$. We found $n=11$ and $m=7$. $m+n = 7+11 = 18$

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Common Mistakes & Tips

• Sign Errors: Be extremely careful with negative signs during algebraic expansion, especially when distributing a negative sign into a bracket, e.g., $-(3n+3) = -3n-3$.
• Memorize Standard Formulas: The variance of the first $N$ natural numbers ($\frac{N^2-1}{12}$) is a common result. Memorizing it can save significant time.
• Leverage Properties: Always look for opportunities to use variance properties like $\text{Var}(aX+b) = a^2 \text{Var}(X)$. This often leads to a much quicker and less error-prone solution.
• Domain Check: Remember that $n$ and $m$ represent counts of numbers, so they must be positive integers.

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Summary

This problem required us to calculate the variance of two different sequences: the first $n$ natural numbers and the first $m$ even natural numbers. We used the fundamental variance formula $\sigma^2 = E[X^2] - (E[X])^2$ along with standard summation formulas to find $n=11$. For $m$, we demonstrated both direct calculation and a more efficient method using the property that $\text{Var}(aX+b) = a^2 \text{Var}(X)$, which yielded $m=7$. Finally, we calculated their sum.

The final answer is $\boxed{18}$.