Sn = [n(n+1)(2n+1)]/ 6

iii.	Sn	=	n(n+1)(2n+1)
			6
Sol.	Sn	=	n(n+1)(2n+1)
			6
∴	S1	=	1(1+1)[2(1)+1]	=	1(2)(3)	=	6	=	1
			6		6		6
∴	S2	=	2(2+1)[2(2)+1]	=	2(3)(5)	=	30	=	5
			6		6		6
∴	S3	=	3(3+1)[2(3)+1]	=	3(4)(7)	=	2× 7	=	14
			6		6

We know that,

t₁ = S₁ = 1

t₂ = S₂ – S₁ = 5 – 1 = 4

t₃ = S₃ – S₂ = 14 – 5 = 9

∴ The first three terms of the sequence are 1, 4 and 9.