= a^n + na^(n – 1)b + [n(n – 1) / 2]a^(n – 2)b^2 + …

Therefore:

a^n = 729 …(1)

na^(n – 1)b

= n (a^n / a) b

= n (729 / a) b

= 729nb / a

Therefore:

729nb/a = 2916

nb / a = 4 ……(2)

[n(n – 1) / 2]a^(n – 2)b^2

= [n(n – 1) / 2] (a^n / a^2) b^2

= 729n(n – 1)b^2 / 2a^2

Therefore:

[729n(n – 1)/2] (b/a)^2 = 4860 ….(3)

From (2):

b/a = 4/n

Substituting for b/a in (3):

729n(n – 1)/2 * 16/n^2 = 4860

5832n(n – 1) = 4860n^2

6n(n – 1) = 5n^2

6n^2 – 6n – 5n^2 = 0

n^2 – 6n = 0

n(n – 6) = 0

Discarding n = 0:

n = 6

Substituting for n in (1):

a = 3

From (2):

b = 4a / n

b = 2

Hence:

a = 3, b = 2, n = 6.

sorry but that is all i can help with the other one is by using pascals triangle