The coefficients of 2nd, 3rd and 4th terms in the expansion of $(1 + x)^{2 n}$ are in A.P., show that $2 n^{2} - 9 n + 7 = 0.$

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Solution

We have,

$(1 + x)^{2 n}$

Now,

Coefficient 2nd term $=^{2 n} C_{2 - 1} =^{2 n} C_{1}$

Coefficient 3rd term $=^{2 n} C_{3 - 1} =^{2 n} C_{2}$

and, Coefficient 4th term $=^{2 n} C_{4 - 1} =^{2 n} C_{3}$

It is given that coefficients are in A.P.

$∴ 2^{2 n} C_{2} =^{2 n} C_{1} +^{2 n} C_{3}$

$\Rightarrow 2 = \frac{^{2 n} C_{1}}{^{2 n} C_{2}} + \frac{^{2 n} C_{3}}{^{2 n} C_{2}}$

$\Rightarrow 2 = \frac{2}{2 n - 2 + 1} + \frac{2 n - 3 + 1}{3}$

$[∵ \frac{^{n} C_{r}}{^{n} C_{r - 1}} = \frac{n - r + 1}{r}]$

$\Rightarrow 2 = \frac{2}{2 n - 1} + \frac{2 n - 2}{3}$

$\Rightarrow 2 = \frac{6 + (2 n - 1) (2 n - 2)}{3 (2 n - 1)}$

$\Rightarrow 6 (2 n - 1) = 6 + 4 n^{2} - 4 n - 2 n + 2$

$\Rightarrow 12 n - 6 = 8 + 4 n^{2} - 6 n$

$\Rightarrow 4 n^{2} - 6 n - 12 n + 8 + 6 = 0$

$\Rightarrow 4 n^{2} - 18 n + 14 = 0$

$\Rightarrow 2 (2 n^{2} - 9 n + 7) = 0$

$\Rightarrow 2 n^{2} - 9 n + 7 = 0$

Hence proved.

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