Q.Find $\displaystyle\sum_{r=1}^{n}(3r^2-2r+1)$.

Beyond a plain A.P. or G.P., many sequences are built from sums or products of simpler patterns, and can be summed using three standard results together with the linearity of summation: the sum of the first $n$ natural numbers is $\sum r=\dfrac{n(n+1)}2$, the sum of their squares is $\sum r^2=\dfrac{n(n+1)(2n+1)}6$, and the sum of their cubes is $\sum r^3=\left[\dfrac{n(n+1)}2\right]^2$. Summation is linear -- $\sum(a_r+b_r)=\sum a_r+\sum b_r$ and $\sum k\,t_r=k\sum t_r$ for a constant $k$, and $\sum 1$ ($n$ times) $=n$ -- so a complicated-looking series is handled by first finding the algebraic expression for its general term $t_r$ (often a polynomial in $r$, built from two arithmetic 'factor' sequences multiplied together), expanding it into separate powers of $r$, and then applying the three results term by term before recombining and simplifying the resulting polynomial in $n$.