Find a, b and n in the expansion of `(a+b)^n`if the first three terms of the expansion are 729, 7290 and 30375, respectively.

Explanation

The first three term of the expansion are given as `729`, `7290`, and `30375`, respectively.
Therefore, we get
`T_(1) = .^(n)C_(0)a^(n-0)b^(0) = a^(n) = 729 " "(1)`
`T_(2) = .^(n)C_(1)a^(n-1)b^(1) = na^(n-1) b = 7290 " " (2)`
`T_(3) = .^(n)C_(2)a^(n-2)b^(2) = (n(n-1))/(2) a^(n-2) b^(2) = 30375" " (3)`
Dividing (2) by (1) , we get
`(na^(n-1)b)/(a^(n)) = (7290)/(729)`
or `(nb)/(a) = 10 " " (4)`
Dividing (3) by (2), we get
`(n(n-1)a^(n-2)b^(2))/( 2na^(n-1)b) = 30375/(7990)`
or `((n-1)b)/(a) = 25/3 " " (5)`
Dividing (4) by (5), we get
`(n)/(n-1) = 6/5`
or `n= 6`
Substituting `n = 6` in (1), we get
`a^(6) = 729` or `a = 3`
From (4), we get
`(6b)/(3) = 10` osr `b = 5`
Thus, `a = 3, b = 5` and `n = 6`.