Let S_K=(1+2+…+K)/K and Σj=1ⁿ Sⱼ²=n/A(B n²+C n+D), where A, B, C, D ∈ ℕ and A has least value. Then

Solution

Maths sir · 4 steps

First simplify S_K to (K+1)/2. Then it is just a sum of squares, shifted by one.

1. Simplify the term. 1+2+…+K = K(K+1)/2, so S_K = (K+1)/2 and Sⱼ² = (j+1)²/4.
2. Shift the index. Put m = j+1. As j runs from 1 to n, m runs from 2 to n+1. That is the full sum of squares up to n+1, minus the missing 1².

   Σⱼ₌₁ⁿ Sⱼ² = 1/4[(n+1)(n+2)(2n+3)/6 - 1] = (2n³+9n²+13n)/24 = n/24(2n²+9n+13)

3. Read off constants. A = 24, B = 2, C = 9, D = 13. The numbers 2, 9 and 13 share no common factor, so 24 is the least possible A. Check with n = 1: 1/24(2+9+13) = 1 = S₁².

   Option	Compute	True?
   (a)	A+B+C+D = 48	no, 48 is not a multiple of 5
   (b)	A+C+D = 46, B = 2	no, 46 is divisible by 2
   (c)	A+B = 26, 5(D-C) = 20	no
   (d)	A+B = 26, D = 13	yes, 26 = 2 × 13

4. So. Only (d) holds: A+B = 26 is divisible by D = 13.

The idea to keep. Always simplify the general term to a clean expression in K first. When the squares start at (j+1)², write the full sum of squares and subtract the terms that are missing.

Answer from the NTA final answer key. Solution by PrepWiser.