Find the 6th term of the expansion `(y^(1//2) + x^(1//3))^(n)` , if the binomial coefficient of 3rd term from the end is 45.

Answer

`252y^(5//2)x^(5//3)`

Explanation

Coefficient of 3rd term from the beginning is always equal to the coefficient of 3rd term frm the end.
Now, `T_(r+1) - .^(n) C(y^(1//2))^(n-r)*(x^(1//3))r.`
Coefficient of `T_(3) = .^(n)C_(2) = (n(n-1))/2.`
` (n(n-1))/2 = 45 rArr n^(2) - n-90 = 0 rArr n^(2) - 10n + 9n - 90 = 0`
`rArr ( n-10 (n+9) = 0 rArr n = 10`
`:. T_(6) = T_(5+1) = .^(10)C_(5) (y^(1//2))^((10-5))*(x^(1//3))^(5)`
`= (10 xx 9 xx 8 xx 7 xx 6)/(5 xx 4 xx 3 xx 2 xx 1) xxy^(5//2)x^(5//3) = 252y^(5//2)x^(5//3).`