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Sum of n Terms

If P2n-1,n:...

Question

If $P (2 n - 1, n) : P (2 n + 1, n - 1) = 22 : 7$ find $n$ .

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Solution

$P (2 n - 1, n) : P (2 n + 1, n - 1) = 22 : 7$

$\frac{P (2 n - 1, n)}{P (2 n + 1, n - 1)} = \frac{22}{7}$

$\frac{\frac{(2 n - 1)!}{{(2 n - 1) - n}!}}{\frac{(2 n + 1)!}{(2 n + 1 - n + 1)!}} = \frac{22}{7}$

$\frac{(2 n - 1)!}{(n - 1)!} \times \frac{(n + 2)!}{(2 n + 1)!} = \frac{22}{7}$

$\frac{(n + 2) (n + 1)}{2 (2 n + 1)} = \frac{22}{7}$

$7 n^{2} + 21 n + 14 = 88 n + 44$

$7 n^{2} - 67 n - 30 = 0$

$(n - 10) (7 n + 3) = 0$

$n = 10$

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