If $P_{r} = (n - r) (n - r + 1) (n - r + 2) . . . . . (n - r + p - 1),$
$Q_{r} = r (r + 1) (r + 2) . . . . (r + q - 1)$ , show that $P_{1} Q_{1} + P_{2} Q_{2} + P_{3} Q_{3} + . . . . . + P_{n - 1} Q_{n - 1} = \frac{⌊ p ⌊ q ⌊ n - 1 + p + q}{⌊ p + q + 1 ⌊ n - 2}$

Open in App

Solution

$\Rightarrow P_{r} = p! \times$ coefficient of $x^{n - r - 1}$ in ${(1 - x)}^{- (p + 1)}$
$Q_{r} = q! \times$ co efficient of $x^{r - 1}$ in ${(1 - x)}^{- (q + 1)}$
Hence, ${(1 - x)}^{- (q + 1)} = \frac{1}{q!} {Q_{1} + Q_{2} x + Q_{3} x^{2} + . . . . + Q_{n - 1} x^{n - 2} + . . . .}$
${(1 - x)}^{- (p + 1)} = \frac{1}{p!} {P_{n - 1} + P_{n - 2} x + P_{n - 3} x^{2} + . . . . + P_{1} x^{n - 2} + . . . .}$
$∴ \frac{S}{p! q!} =$ the co efficient of $x^{n - 2}$ in ${(1 - x)}^{- (p + q + 2)}$
$= \frac{(p + q + n - 1)!}{(p + q + 1)! (n - 2)!}$
$S = \frac{p! q! (p + q + n - 1)}{(p + q + 1)! (n - 2)!}$

Suggest Corrections

Similar questions

\sum_{r = 0}^{n} r^{2} c_{r} p^{r} \cdot q^{n - r} = n p q + n^{2} p^{2}

^{n} C_{0},^{n} C_{1},^{n} C_{2}, \dots,^{n} C_{n}

(1 + x)^{n}

p + q = 1

n \sum r = 0 r^{2}^{n} C_{r} p^{r} q^{n - r}

\sum_{r = 0}^{n} r^{2} .^{n} C_{r} p^{r} q^{n - r}, w h e r e p + q = 1,

0, 1, 2, \dots, n

q^{n},^{n} C_{r} q^{n - 1} p,^{n} C_{2} q^{n - 2} p^{2}, \dots, p^{n}

p + q = 1,

n^{2} p^{2} + n p q

n p q .

Join BYJU'S Learning Program