If n > 2, the prove that
$C_{1} (a - 1)^{2} - C_{2} (a - 2)^{2} + . . . + (- 1)^{n - 1} C_{n} (a - n)^{2} = a^{2}$

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Solution

(a) $T r = (- 1)^{r - 1} C_{r} (a - r)^{2}$
$∴ S_{n} = \sum_{r = 1}^{n} T_{r} = \sum_{r = 1}^{n} (- 1)^{r - 1} C_{r} (a^{2} - 2 a r + r^{2})$
$= (- 1)^{r - 1} {a^{2} \sum_{r = 1}^{n} C_{r} - 2 a \sum_{r = 1}^{n} r C_{r}, + a^{2} \sum_{r = 1}^{n} r^{2} C_{r}}$
The last two sums are zeros by Q.3 (a) and (b) if we put $x = - 1$ because of alt. +ive and - ive signs.
The first sum
$C_{1} - C_{2} + C_{3} - C_{4} + . . . . . = C_{0} = 1$
$∴ S_{n} = a^{2} . 1 + 0 + 0 = a^{2}$

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