Q.Find the mean of first n natural numbers.

Mean and Variance of Natural Numbers

Let’s start with something you already know: the mean (average) and variance (spread) of a set of numbers. If I give you the first five natural numbers — 1, 2, 3, 4, 5 — you can compute their mean and variance easily. But what if I ask: What is the mean of all natural numbers? That’s infinite, so it doesn’t make sense directly. Instead, we ask: What is the mean of the first $n$ natural numbers? And then we see how it behaves as $n$ grows.

That’s the core idea: we study the mean and variance of the first $n$ natural numbers as a function of $n$, and often look at what happens when $n$ becomes very large.

Intuition First

Imagine you line up the numbers $1, 2, 3, \dots, n$ on a number line. Their average is somewhere in the middle — roughly $n/2$. More precisely, the mean of the first $n$ natural numbers is $\frac{n+1}{2}$. For $n=5$, that’s $3$, which matches your intuition.

Now, variance measures how spread out the numbers are around that mean. For small $n$, the spread is small; for large $n$, the spread grows. The variance of the first $n$ natural numbers turns out to be $\frac{n^2 - 1}{12}$. For $n=5$, that’s $\frac{25-1}{12} = 2$, which is a moderate spread.

Precise Statement

Let $X$ be a random variable that takes values $1, 2, 3, \dots, n$ with equal probability $1/n$. Then:

These are exact formulas for any positive integer $n$.

Derivation (Why These Formulas?)

Mean

The sum of the first $n$ natural numbers is $1 + 2 + \dots + n = \frac{n(n+1)}{2}$.

Since there are $n$ numbers, the mean is:

Variance

Variance is the average of squared deviations from the mean:

A cleaner way uses the identity: $\sigma^2 = E[X^2] - (E[X])^2$.

First, $E[X^2] = \frac{1}{n} \sum_{k=1}^n k^2$. The sum of squares formula is $\sum_{k=1}^n k^2 = \frac{n(n+1)(2n+1)}{6}$. So:

Now, $(E[X])^2 = \left( \frac{n+1}{2} \right)^2 = \frac{(n+1)^2}{4}$.

Therefore:

Factor $(n+1)$:

Compute the bracket: common denominator 12:

Thus:

What This Tells You

• The mean grows linearly with $n$ — roughly half of $n$.
• The variance grows quadratically — roughly $n^2/12$ for large $n$.
• For large $n$, the standard deviation $\sigma_n \approx \frac{n}{\sqrt{12}} \approx 0.2887n$, meaning the spread is about 29% of the range. …