2.3 Average of Squares and Cubes of First n Natural Numbers
Free study material · concepts, shortcuts & solved questions
(a) Sum of squares of first n natural numbers: n^2 = (n(n+1)(2n+1))/(6) So, Average of squares of first n natural numbers = ((n+1)(2n+1))/(6)
(b) Sum of cubes of first n natural numbers: n^3 = [(n(n+1))/(2)]^2 So, Average of cubes of first n natural numbers = (n(n+1)^2)/(4)
Solved Example 7: Find the average of the squares of the first 10 natural numbers.
Average = (n+1)(2n+1)/6 = (11)(21)/6 = 231/6 = 38.5.
Solved Example 8: Find the average of the cubes of the first 9 natural numbers.
Average = n(n+1)²/4 = 9 × 100/4 = 900/4 = 225.
Solved Example 9: Find the average of the squares of the first 12 natural numbers.
Average = (13)(25)/6 = 325/6 = 54.17 (approx).
Note on derivation (for conceptual clarity, not required to memorise): The sum-of-squares formula n(n+1)(2n+1)/6 and sum-of-cubes formula [n(n+1)/2]² are both standard results proven by mathematical induction. You are never expected to derive them in the exam — only to apply them correctly and quickly. A useful memory hook: the sum of cubes of the first n natural numbers always equals the square of the sum of the first n natural numbers. That is, n^3 = ( n)^2. This is a beautiful and exam-useful identity — for instance, sum of cubes of first 5 natural numbers = (1+2+3+4+5)² = 15² = 225, and indeed 1+8+27+64+125 = 225. ✓
Solved Example 9(a): Find the average of the squares of the first 15 natural numbers.
Average = (n+1)(2n+1)/6 = (16)(31)/6 = 496/6 = 82.67 (approx).
Solved Example 9(b): Find the average of the squares of the numbers from 6 to 10 (inclusive).
This is a crucial edge case: the standard “first n natural numbers” formula does NOT apply here directly, because the numbers do not start from 1. The correct method is to compute the sum of squares directly (or subtract the sum-of-squares up to 5 from the sum-of-squares up to 10) and divide by the actual count. Squares: 6²=36, 7²=49, 8²=64, 9²=81, 10²=100. Sum = 36+49+64+81+100 = 330. Count = 5. Average = 330/5 = 66. (Cross-check using the subtraction method: sum of squares from 1 to 10 = 10×11×21/6 = 385; sum of squares from 1 to 5 = 5×6×11/6 = 55; sum from 6 to 10 = 385−55 = 330. ✓ Same answer — this confirms why blindly plugging n=5 into the “first n” formula would have been wrong.)