If s and t respectively the sum and the sum of the squares of n successive positive integers beginning with a, then show that $n t - s^{2}$ is independent of a.

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Solution

$a, a + 1, a + 2 + . . . . . . + (a + n - 1)$
$s = n a + \frac{n (n - 1)}{2} . . . (1)$
$t = a^{2}, (a + 1)^{2}, + . . . . . . + (a + n - 1)^{2}$
$= n a^{2} + 2 a \cdot \frac{n (n - 1)}{2} + \sum_{N = 1}^{n - 1} N^{2}$
or $n t = n^{2} a^{2} + a n^{2} (n - 1) + n \sum N^{2}$
$a n d s^{2} = n^{2} a^{2} + a n^{2} (n - 1) + \frac{n^{2} (n - 1)^{2}}{4}, b y (1)$
$n t - s^{2}$ is clearly independent of a

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