The difference between the sum of the cubes of first n natural numbers and squares of first n natural numbers is

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Solution

The correct option is B

The sum of cubes of first n natural numbers = $\frac{n^{2} (n + 1)^{2}}{4}$

The sum of squares of first n natural numbers = $\frac{n (n + 1) (2 n + 1)}{6}$

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The remainder obtained when the sum of squares first n odd natural numbers is divided the sum of squares of first n even natural numbers is

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