Question -
Answer -
Let Tn bethe nth term of the given series.
We have:
Tn =[1 + (n – 1)2]3
= (2n – 1)3
= (2n)3 –3 (2n)2. 1 + 3.12.2n-13 [Since, (a-b)3 =a3 – 3a2b + 3ab2 – b]
= 8n3 –12n2 + 6n – 1
Now, let Sn bethe sum of n terms of the given series.
We have:
Upon simplification weget,
= 2n2 (n+ 1)2 – n – 2n (n + 1) (2n + 1) + 3n (n + 1)
= n (n + 1) [2n (n +1) – 2 (2n + 1) + 3] – n
= n (n + 1) [2n2 –2n + 1] – n
= n [2n3 –2n2 + n + 2n2 – 2n + 1 – 1]
= n [2n3 –n]
= n2 [2n2 –1]
∴ The sum of the seriesis n2 [2n2 – 1]