Question -
Answer -
Let Tn bethe nth term and Sn be the sum to n terms of the givenseries.
We have,
Sn = 3+ 5 + 9 + 15 + 23 + …………. + Tn-1 + Tn … (1)
Equation (1) can berewritten as:
Sn = 3+ 5 + 9 + 15 + 23 + …………. + Tn-1 + Tn ……..(2)
By subtracting (2)from (1) we get
Sn = 3+ 5 + 9 + 15 + 23 + …………. + Tn-1 + Tn
Sn = 3+ 5 + 9 + 15 + 23 + …………. + Tn-1 + Tn
0 = 3 + [2 + 4 + 6 + 8+ … + (Tn – Tn-1)] – Tn
The difference betweenthe successive terms are 5-3 = 2, 9-5 = 4, 15-9 = 6,
So these differencesare in A.P
Now,
∴ The sum of the seriesis n/3 (n2 + 8)