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Binomial Expression

If the coeffi...

Question

If the coefficients of (r - 1)th , rth and (r + 1)th terms in the expansion of (1 + x) $^{n}$ be in H.P then prove that n is a root of the equation $x^{2} - (4 r - 1) x + 4 r^{2} = 0$

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Solution

$2 \frac{1}{^{n} C_{r}} = \frac{1}{^{n} C_{r - 1}} + \frac{1}{^{n} C_{r + 1}}$

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

Solving thia gives us

$n^{2} + 4 r^{2} - 4 n r + n = 0$

$∴$ n is a root of the equation.

$x^{2} - (4 r - 1) x + 4 r^{2} = 0$

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