Question:medium

The average of the squares of the first 50 natural numbers is

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The sum of squares of the first \( n \) natural numbers can be calculated using the formula \( \dfrac{n(n + 1)(2n + 1)}{6} \), and the average is obtained by dividing the sum by \( n \).
Updated On: May 7, 2026
  • \( \dfrac{1817}{2} \)
  • \( \dfrac{1717}{2} \)
  • \( \dfrac{1517}{2} \)
  • \( \dfrac{1617}{2} \)
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The Correct Option is B

Solution and Explanation

The sum of the squares of the first $n$ natural numbers is given by the formula:
Sum = $\frac{n(n+1)(2n+1)}{6}$.
The average of these squares is the sum divided by the number of terms, $n$.
Average = $\frac{\text{Sum}}{n} = \frac{n(n+1)(2n+1)}{6n} = \frac{(n+1)(2n+1)}{6}$.
In this problem, we need to find the average for the first 50 natural numbers, so $n=50$.
Substitute $n=50$ into the average formula:
Average = $\frac{(50+1)(2 \times 50+1)}{6}$.
Average = $\frac{(51)(100+1)}{6}$.
Average = $\frac{51 \times 101}{6}$.
We can simplify the fraction by dividing 51 and 6 by their common factor, 3.
Average = $\frac{17 \times 101}{2}$.
Now, calculate the product in the numerator: $17 \times 101 = 1717$.
So, the average is $\frac{1717}{2}$.
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