Find the sum of the first n terms of the series: 3 + 7 + 13 + 21 + 31 + …
This given series is neither an AP nor GP.
Let
Sn = 3 + 7 + 13 + 21 + 31 + … + an - 1 + an …(1)
Sn = 0 + 3 + 7 + 13 + 21 + 31 + … + an - 2 + an - 1 + an …(2)
Subtracting (2) from (1)
Sn - Sn = (3 - 0) + [(7 - 3) + (13 - 7) + (21 - 13) + … + (an - 1 - an - 2)
+ (an - an - 1) - an
⇒ 0 = 3 + [4 + 6 + 8 + … an - 1] - an
⇒ an = 3 + [4 + 6 + 8 + … an - 1] …(3)
Now, 4 + 6 + 8 + … an - 1 is an AP.
Where,
first term(a) = 4
common difference(d) = 6 - 4 = 2
We know that,
Sum of n terms of AP = (n/2)[2a + (n - 1)d]
putting n = n - 1, a = 4, d =2
[4 + 6 + 8 + … to (n - 1) terms] = (n - 1)/2 × [2a + (n - 1 - 1)d]
= (n - 1)/2 × [2(4) + (n - 2)2]
= (n - 1)/2 × [8 + 2n - 4]
= (n - 1)/2 × [2n + 4]
= (n - 1)/2 × 2(n + 2)
= (n - 1)(n + 2)
∴
an = 3 + [4 + 6 + 8 + … an - 1]
= 3 + (n - 1)(n + 2)
= 3 + n2 + 2n - n - 2
= 3 + n2 + n - 2
= n2 + n + 1
Now,
Sn
Thus, the required sum is .