Question

if the coefficients of 2nd, 3rd and 4th terms in the expansion of (1 + x)^2n are in AP show that 2n² - 9n + 7= 0​

Answer 1

Answer:

Given , the expansion

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

The General term for the above expansion is

$T_{r + 1} = \: ^{2n}C_{r} {1}^{2n - r} {x}^{r}$

So the 2nd term is

$T_{2} = {}^{2n}C_{1} {x}^{1} \\ \implies T_{2} = 2nx$

The 3rd term is

$T_{3} = {}^{2n} T_2 {x}^{2} \\ \implies \: T_3 = \frac{ 2n(2n - 1) {x}^{2} }{2}$

And the 4th term is

$T_4 = {}^{2n} C_3 {x}^{3} \\ \implies \: T_4 = \frac{2n(2n - 1)(2n - 2)}{6}$

According to question the coefficients of

$T_2 \: \: T_3 \: \: and \: T_4$

are in AP

So

$\implies2 \times \frac{2n(2n - 1) }{2} = 2n + \frac{2n(2n - 1)(2n - 2)}{6} \\ \implies {(2n - 1)}= 1 + \frac{(2n - 1)(2n - 2)}{6} \\ \implies (2n - 1) = \frac{6 + (2n - 1)(2n - 2)}{6} \\ \implies6(2n - 1) = 6 + (2n - 1)(2n -2 ) \\ \implies3(2n - 1) = 3 + (2n - 1)(n - 1) \\ \implies6 {n}^{2} - 3 = 3 + 2 {n}^{2} - 3n + 1 \\ \implies2 {n}^{2} - 9n + 7 = 0$