Find the sum of the first n terms of the series: 3 + 7 + 13 + 21 + 31 + …

The given series is 3 + 7 + 13 + 21 + 31 + …

S = 3 + 7 + 13 + 21 + 31 + …+ a_n–1 + a_n

S = 3 + 7 + 13 + 21 + …. + a_{n – 2} + a_{n – 1} + a_n

On subtracting both the equations, we get

S – S = [3 + (7 + 13 + 21 + 31 + …+ a_n–1 + a_n)] – [(3 + 7 + 13 + 21 + 31 + …+ a_n–1) + a_n]

S – S = 3 + [(7 – 3) + (13 – 7) + (21 – 13) + … + (a_n – a_n–1)] – a_n

0 = 3 + [4 + 6 + 8 + … (n –1) terms] – a_n

a_n = 3 + [4 + 6 + 8 + … (n –1) terms]

\(=3 +(\dfrac{n-1}{2})[2\times 4+(n-1-1)2]\)

= \(3 +(\dfrac{n-1}{2})[8+(n-2)2]\)

= \(3 +(\dfrac{n-1}{2})(2n+4)\)

\( =3+(n-1)(n+2)\)

\( =3+(n^2+n-2)\)

\( =n^2+n+1\)

Therefore,

\( \displaystyle\sum_{k=1}^{n} a_k=\displaystyle\sum_{k=1}^{n}k^2+\displaystyle\sum_{k=1}^{n} k+\displaystyle\sum_{k=1}^{n}1\)

\( \dfrac{n(n+1)(2n+1)}{6}+\dfrac{n(n+1)}{2}+n\)

= \( n[ \dfrac{(n+1)(2n+1)+3(n+1)+6}{6}]\)

= \( n[ \dfrac{2n^2+3n+13n+3+36}{6}]\)

= \( n[ \dfrac{2n^2+6n+10}{6}]\)

= \( \dfrac{n}{3}[n^2+3n+5]\)

Therefore, the sum of n terms of the given series

\( \dfrac{n}{3}[n^2+3n+5]\)