Find the sum of the squares of the first twelve positive multiples of 8.
Solution
✅ Correct Option: 3
The first twelve positive multiples of 8 are $8,16,\ldots,96$. Their sum of squares is $8^2(1^2+2^2+\cdots+12^2)$. Using $1^2+2^2+\cdots+n^2=\dfrac{n(n+1)(2n+1)}{6}$: $64\times\dfrac{12\times13\times25}{6}=64\times650=41600$.
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