(v(n + 1))ⁿ+¹ = (v(n+1))ⁿ x v(n + 1)
= u1 u2 u3...un u(n+1)

(vn)ⁿ = u1 u2 u3...un


(v(n + 1))ⁿ+¹ ÷ (vn)ⁿ = (u1 u2 u3...un u(n +1)) ÷ (u1 u2 u3... un)


(v(n+1))ⁿ x v(n + 1) ÷ (vn)ⁿ = u(n+1)

v(n+1))ⁿ / (vn)ⁿ = u(n+1) / v(n+1)


( v(n+1) / vn )ⁿ = arⁿ / v(n+1)

( v(n+1) / vn )ⁿ = arⁿ / (a ar ar² ar³...arⁿ)¹/ⁿ+¹

= arⁿ / [a(ⁿ+¹)/(ⁿ+¹) × rⁿ(ⁿ+¹)/²/(ⁿ+¹) ]

= arⁿ / (a¹ x rⁿ/²)

= rⁿ/²

So,

(v(n+1) / vn ) = (rⁿ/²)¹/ⁿ

= r¹/²

= √r (constant)


Example : Let a = 2, r = 3,

2 x 2 x 3¹ x 2 x 3² x 2 x 3³ = 2⁴ x 3^6 = (v4)⁴
v4 = 2 x 3³/²

2 x 2 x 3¹ x 2 x 3² = 2³ x 3³ = (v3)³
v3 = 2 x 3¹

v4 / v3 = 2 x 3³/² ÷ (2 x 3¹)
= 3¹/²

= √3

Alternative for i) :


vn
= (u1 u2 u3...un)¹/ⁿ
= (a ar ar² ar³ ... arⁿ-¹)¹/ⁿ
= ( aⁿ x r(¹+ⁿ-¹)(ⁿ-¹)/² )¹/ⁿ

= ( aⁿ x rⁿ(ⁿ-¹)/² )¹/ⁿ

= a r(ⁿ-¹)/²

So,

v(n+1) = a r(ⁿ+¹ - ¹)/² = a rⁿ/²


Therefore , v(n+1) / vn

= a rⁿ/² ÷ a r(ⁿ-¹)/²

= rⁿ/² - (ⁿ-¹)/²

= r(ⁿ - ⁿ+¹)/²

= r¹/²

= √r