Find the $n^{th}$ term and deduce the sum to n terms of the series $4+11+22+37+56+\cdots$
$
\begin{aligned}
\text{Let } S_n & = & 4 + & 11 + 22 + 37 + 56 + \cdots + t_{n-1} + t_n \\
& & & \text{Shifting every term one place to the right} \\
S_n & = & & 4 + 11 + 22 + 37 + \cdots \cdots \cdots + t_{n-1} + t_n
\end{aligned}
$
Subtracting, we get
$
\begin{aligned}
0 & = \left[4 + 7 + 11 + 15 + 19 + \cdots \text{to n terms}\right] - t_n \\
\text{i.e. } t_n & = 4 + \left[7 + 11 + 15 + 19 + \cdots \text{to } \left(n-1\right) \text{ terms}\right] \\
& = 4 + \left(\dfrac{n-1}{2}\right)\left[2 \times 7 + \left(n-2\right) \times 4\right] \\
& \left[\text{Note: Sum to n terms of an A.P } = \dfrac{n}{2}\left(2a+\left(n-1\right)d\right)\right] \\
& \left[\text{Here, first term }= a = 7 \; ; \text{common difference }= d = 4 \right] \\
& = 4 + \dfrac{\left(n-1\right) \left(4n+6\right)}{2} \\
& = 4 + \dfrac{4n^2 + 2n - 6}{2} \\
& = 4 + 2n^2 + n - 3 \\
\text{i.e. } t_n & = 2n^2 + n + 1
\end{aligned}
$
$
\begin{aligned}
\therefore \text{ Sum to n terms} = S_n & = 2 \sum_{k = 1}^{n} k^2 + \sum_{k=1}^{n} k + n \\
& = 2 \times \dfrac{n \left(n+1\right) \left(2n+1\right)}{6} + \dfrac{n \left(n+1\right)}{2} + n \\
& = \dfrac{n}{3} \left(2n^2 + 3n + 1\right) + \dfrac{n \left(n+1\right)}{2} + n \\
& = \dfrac{n}{6} \left(4n^2 + 6n + 2 + 3n + 3 + 6\right) \\
\therefore S_n & = \dfrac{n}{6} \left(4n^2 + 9n + 11\right)
\end{aligned}
$