(iii)
From previous part we know that Un = 4/3 (3ⁿ) - 2n - 1 So likewise, Ur = 4/3 (3^r) - 2r - 1
Starting from r = 1, for a total of n terms,
∑ Ur
= ∑ ( 4/3 (3^r) - 2r - 1)
= ∑ 4/3 (3^r) - ∑ 2r - ∑ 1
= 4/3 ∑ (3^r) - 2∑ r - ∑ 1
Now, 3^r is our usual geometric progression (G.P), with a common ratio r = 3 and first term a = 3¹ = 3
So we can use our sum of GP formula Sn = a(rⁿ - 1)/(r - 1) = 3(3ⁿ - 1)/(3 - 1) = 3(3ⁿ - 1) / 2
∑r is the sum of n consecutive integers, from 1 to n. So we can use the formula ½n(n+1) (this is often called the pairing or rainbow method at O levels/PSLE)
∑1 is simply adding up n '1's together, so it just equals n.
Putting them altogether,
∑ Ur = 4/3 ∑ (3^r) - 2∑r - ∑ 1
= 4/3 (3(3ⁿ - 1)/2) - 2(½n(n+1)) - n
= 2(3ⁿ - 1) - n(n+1) - n
= 2(3ⁿ) - 2 - n² - n - n
= 2(3ⁿ) - n² - 2n - 2