17. If the coefficients 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

It is given that `.^(2n) C_(1), ^(2n) C_(2) and .^(2n) C_(3)` are in AP.
` :. 2 xx .^(2n) C_(2) = .^(2n) C_(1) + .^(2n) C_(3)`
`rArr 2=(.^(2n)C_(1))/(.^(2nC_(2)))+(.^(2n)C_(3))/(.^(2n) C_(2))`
` rArr = 2= (4n)/(2n(2n-1))+(2n-3+1)/3 " " [:' (.^(2n)C_(r))/(.^(2n)C_(r-1))=(2n-r+1)/r and r=3]`
`rArr 2/(2n-1)+(2n-2)/3 = 2`
` rArr 6+ (2n-1) (2n-2) = 6 ( 2n-1)`
`rArr 4n^(2) - 18n + 14 = 0 rArr 2n^(2) - 9n + 7 =0.`