First slide

Binomial theorem for positive integral Index

Question

If the coefficients of P th, ( p + 1)th and (p + 2)th terms in the expansion of (1 + x)ⁿ are in AP, then

Moderate

A

$n^{2} - 2 np + 4 p^{2} = 0$
B

$n^{2} - n (4 p + 1) + 4 p^{2} - 2 = 0$
C

$n^{2} - n (4 p + 1) + 4 p^{2} = 0$
D

None of these above

Solution

coefficients of pth , (p + 1)th and (p + z)th terms in the expansion (1 + x)" are $^{n} C_{p - 1},^{n} C_{p^{'}}^{} n_{C_{p + 1,}}$ respectively.
Since, these are in AP.
$∴ 2^{n} C_{p} =^{n} C_{p - 1} +^{n} C_{p + 1}$
$\Rightarrow 2 \frac{n!}{(n - p)! p!} = \frac{n!}{(n - p + 1)! (p - 1)!} + \frac{n!}{(n - p - 1)! (p + 1)!}$

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

$\Rightarrow \frac{2}{1} = \frac{p}{(n - p + 1)} + \frac{n - p}{p + 1}$
$\Rightarrow n^{2} - n (4 p + 1) + 4 p^{2} - 2 = 0$

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