Jawab:
(n(2n+1)(n+1))/6 ✔️

Penjelasan dengan langkah-langkah:
Bilangan kuadrat {1², 2², 3², 4², ....}

a = 1, b = 4,     9,   16
........+3, c = +5 + 7
d = ... +2 + 2

Maka a = 1, b = 4, c = 5, d = 2

Rumus Sn bertingkat
Sn = a/0! + b(n-1)/1! + c(n-1)(n-2)/2! + d(n-1)(n-2)(n-3)/3!
Sn = 1/0!+ 4(n-1)/1! + 5(n-1)(n-2)/2! + 2(n-1)(n-2)(n-3)/3!
∵ 0! = 1! = 1 ∴
Sn = 1 + 4(n-1) + 5(n-1)(n-2)/2 + 2(n-1)(n-2)(n-3)/(3×2)
Sn = 1 + 4n-4 + 5(n-1)(n-2)/2 + (n-1)(n-2)(n-3)/3
Sn = 4n-3 + 5(n²-2n-n+2)/2 + (n²-2n-n+2)(n-3)/3
Sn = 4n-3 + 5(n²-3n+2)/2 + (n²-3n+2)(n-3)/3
Sn = 4n-3 + (5n²-15n+10)/2 + (n³-3n²+2n-3n²+9n-6)/3
Sn = 4n-3 + (5n²-15n+10)/2 + (n³-6n²+11n-6)/3
Sn = 6(4n-3)/6 + 3(5n²-15n+10)/2(3) + 2(n³-6n²+11n-6)/3(2)
Sn = (24n-18)/6 + (15n²-45n+30)/6 + (2n³-12n²+22n-12)/6
Sn = (24n-18+15n²-45n+30+2n³-12n²+22n-12)/6
Sn = (2n³+15n²-12n²+24n+22n-45n+30-12-18)/6
Sn = (2n³+3n²+n)/6
Sn = (n(2n²+3n+1))/6
Sn = (n((1/2)(2n+1)(2n+2)))/6
Sn = (n(2n+1)(n+1))/6 ✔️

(xcvi)