The average of the squares of the first 45 natural numbers is

Correct — B. Average the squares; do not square the average.

Sum of squares of the first n naturals = n(n + 1)(2n + 1) ⁄ 6
Divide by n and the average is simply (n + 1)(2n + 1) ⁄ 6

n = 45: 46 × 91 = 4,186
4,186 ⁄ 6 = 697.666… = 697.67 → option (b)

The long route agrees: 45 × 46 × 91 ⁄ 6 = 31,395, and 31,395 ⁄ 45 = 697.67.

Why the others are wrong

• (a)
  699.67 — 699.67 sits two above the answer. Multiplied by 6 it needs the product to be 4,198, which is 12 more than the 4,186 that 46 × 91 actually gives.
• (c)
  696.67 — 696.67 is 4,180 ⁄ 6 — the product read six short. Compute it as 46 × 90 + 46 = 4,140 + 46 = 4,186.
• (d)
  698.67 — 698.67 is 4,192 ⁄ 6. Every option here ends in .67, so the decimal tail proves nothing; the item turns entirely on reaching the integer 4,186.

The sum of the squares of the first n natural numbers is n(n + 1)(2n + 1) ⁄ 6. Since there are n of them, the average of the squares is (n + 1)(2n + 1) ⁄ 6.

Dividing by n before multiplying saves the whole large-number step. You never need 31,395 unless you want it as a check.

Note that this is not the square of the average. The first 45 naturals average 23, and 23² = 529, which is far below 697.67 — squaring is not an averaging-friendly operation.

Key facts

• Sum of the squares of the first n naturals = n(n + 1)(2n + 1) ⁄ 6.
• The average of those squares reduces to (n + 1)(2n + 1) ⁄ 6.
• For n = 45 that is 46 × 91 ⁄ 6 = 4,186 ⁄ 6 = 697.666…, which rounds to 697.67.
• The sum itself is 31,395.

Common traps

• Using n(n + 1) ⁄ 2, the sum of the numbers themselves, instead of the sum of their squares
• Dividing the sum of squares by 46 instead of 45
• Squaring the average of the first 45 numbers rather than averaging the squares

No directly related past PYQ was found.