The coefficient of ${(r - 1)}^{t h}, {(r)}^{t h}, a n d {(r + 1)}^{t h}$ terms in the expansion of ${(r - 1)}^{t h}$ are in the ratio $1 : 3 : 5$ . Find $n$ and $r$ .

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Solution

We have,

$\frac{c o e f f i c i e n t o f {(r - 1)}^{t h} t e r m s}{c o e f f i c i e n t o f r^{t h} t e r m s} = \frac{1}{3}$

$\frac{^{n} C_{r - 2}}{^{n} C_{r - 1}} = \frac{1}{3}$

$\frac{[\frac{n!}{(r - 2)! (n - r + 2)!}]}{[\frac{n!}{(r - 1)! (n - r + 1)!}]} = \frac{1}{3}$

$\frac{(r - 1)! (n - r + 1)!}{(r - 2)! (n - r + 2)!} = \frac{1}{3}$

$\frac{(r - 1) (r - 2)! (n - r + 1)!}{(r - 2)! (n - r + 2) (n - r + 1)!} = \frac{1}{3}$

$\frac{(r - 1)}{(n - r + 2)} = \frac{1}{3}$

$3 r - 3 = n - r + 2$

$4 r = n + 5$ ……… (1)

Similarly,

$\frac{c o e f f i c i e n t o f r^{t h} t e r m s}{c o e f f i c i e n t o f {(r + 1)}^{t h} t e r m s} = \frac{3}{5}$

$\frac{^{n} C_{r - 1}}{^{n} C_{r}} = \frac{3}{5}$

$\frac{\frac{n!}{(r - 1)! (n - r + 1)!}}{\frac{n!}{r! (n - r)!}} = \frac{3}{5}$

$\frac{r! (n - r)!}{(r - 1)! (n - r + 1)!} = \frac{3}{5}$

$\frac{r (r - 1)! (n - r)!}{(r - 1)! (n - r + 1) (n - r)!} = \frac{3}{5}$

$\frac{r}{(n - r + 1)} = \frac{3}{5}$

$5 r = 3 n - 3 r + 3$

$8 r = 3 n + 3$ ………. (2)

From equation (1) and (2), we get

$n = 7, r = 3$

Hence, the value of $n, r = 7, 3$ , respectively.

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