Sum the series
$a b C_{0} - (a - 1) (b - 1) C_{1} + (a -, 2) (b - 2) C_{2} + . . . + (- 1)^{n} (a - n) . (b - n) C_{n}$
where a,b are real numbers and $C_{r} =^{n} C_{r}, n > 3.$

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Solution

$L . H . S . = (- 1)^{r} \sum_{r = 0}^{n} (a - r) (b - r) C_{r}$
$= (- 1)^{r} \sum_{r = 0}^{n} [a b (a + b) r + r^{2}] C_{r}$
$= (- 1)^{r} [a b \sum_{r = 0}^{n} C_{r} - (a + b) \sum_{r = 0}^{n} r C_{r} + \sum_{r = 0}^{n} r^{2} C_{r}] = 0$
Because of $(- 1)^{r}$ the terms in each sum are alternately +ive and -ive .
Hence by Q.1, as in Q.3 (b) on putting x = -1 and result (A) p. 389

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Sum the series abC0−(a−1)(b−1)C1+(a−,2)(b−2)C2+...+(−1)n(a−n).(b−n)Cnwhere a,b are real numbers and Cr=nCr,n>3.

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Sum the series
$a b C_{0} - (a - 1) (b - 1) C_{1} + (a -, 2) (b - 2) C_{2} + . . . + (- 1)^{n} (a - n) . (b - n) C_{n}$
where a,b are real numbers and $C_{r} =^{n} C_{r}, n > 3.$