If the coefficient of 2nd, 3rd and 4th terms in the expansion of `(1+x)^(2n)` are in A.P. , show that `2n^2-9n+7=0.`

Explanation

Given expansion is `(1 + x)^(2n)`
Now, coefficient of 2nd term `= .^(2n)C_(1)`
Coefficient of 3rd term `= .^(2n)C_(2)`
Coefficient of 4th term `= .^(2n)C_(3)`
Given that, `.^(2n)C_(1), .^(2n)C_(2) " and " .^(2n)C_(3)` are in AP.
Then, `2. .^(2n)C_(2) = .^(2n)C_(1) + .^(2n)C_(3)`
`rArr 2 [(2n (2n - 1) (2n - 2)!)/(2 xx 1 xx (2n - 2)!)] = (2n (2n -1)!)/((2n - 1)!) + (2n (2n -1) (2n - 2) (2n - 3)!)/(3! (2n - 3)!)`
`rArr n(2n - 1) = n + (n(2n - 1) (2n -2))/(6)`
`rArr n(12n - 6) = n (6 + 4n^(2) - 4n - 2n + 2)`
`rArr 12 n - 6 = (4n^(2) - 6 n + 8)`
`rArr 6(2n - 1) = 2 (2n^(2) - 3n + 4)`
`rArr 3(2n - 1) = 2n^(2) - 3n + 4`
`rArr 2n^(2) - 3n + 4 - 6n + 3 = 0`
`rArr 2n^(2) - 9n + 7 = 0`