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.