Question

Find the sum upto n terms of the series-
$3 + 7 + 13 + 21 + 31 + ...n$ terms

Answer 1

Hint: To find the sum of the series, we will first check if it is an AP or a GP. An AP is a series where successive terms have the same common difference. For example- 1, 3, 5, 7, 9, … A GP is a series where successive terms have the same ratio. For example- 2, 4, 8, 16, 32, …

Complete step-by-step answer:
In the given series $3 + 7 + 13 + 21 +31..,$ we can clearly see that it is neither an AP nor a GP. The common difference and ratio keep on changing as-
$7 - 3 \ne 13 - 7$
$\dfrac{7}{3} \ne \dfrac{{13}}{7}$
When we look closely, the common difference of successive terms are
$7 - 3 = 4$
$13 - 7 = 6$
$21 - 13 = 8$
$31 - 21 = 10$
We can see that the common differences of successive terms are forming an AP. These types of series can be solved by method of difference as-
$\begin{align}
  &{{\text{S}}_{\text{n}}} = 3 + 7 + 13 + 21 + .... + {{\text{a}}_{{\text{n}} - 1}} + {{\text{a}}_{\text{n}}} \\
  &{{\text{S}}_{\text{n}}} = 0 + 3 + \;7\; + 13 + .... + {{\text{a}}_{{\text{n}} - 2}} + {{\text{a}}_{{\text{n}} - 1}} + {{\text{a}}_{\text{n}}} \\
  \text{ Subtracting term by term we get -} \\
  &0 = 3 + \left[ {\left( {7 - 3} \right) + \left( {13 - 7} \right) + \left( {21 - 13} \right) + ...\left( {{{\text{a}}_{\text{n}}} - {{\text{a}}_{{\text{n}} - 1}}} \right)} \right] - {{\text{a}}_{\text{n}}} \\
  &{{\text{a}}_{\text{n}}} = 3 + \left[ {4 + 6 + 8 + ... + {{\text{a}}_{{\text{n}} - 1}}} \right] \\
\end{align} $
The terms in the bracket form an AP of (n - 1) terms with first term 4 and common difference 2. The sum of an AP is given by-
${\text{S}} = \dfrac{{\text{n}}}{2}\left( {2{\text{a}} + \left( {{\text{n}} - 1} \right){\text{d}}} \right)$
Putting $n = n - 1$, $a = 4$ and $d = 2$,
$\begin{align}
  &{\text{S}} = \dfrac{{{\text{n}} - 1}}{2}\left( {2 \times 4 + \left( {{\text{n}} - 2} \right)2} \right) \\
  &{\text{S}} = \left( {{\text{n}} - 1} \right)\dfrac{{\left( {2{\text{n}} - 4 + 8} \right)}}{2} = \left( {{\text{n}} - 1} \right)\dfrac{{\left( {2{\text{n}} + 4} \right)}}{2} \\
  &{\text{S}} = \left( {{\text{n}} - 1} \right)\left( {{\text{n}} + 2} \right) \\
\end{align} $
So we can substitute the value of S as-
$\begin{align}
  &{{\text{a}}_{\text{n}}} = 3 + \left( {{\text{n}} - 1} \right)\left( {{\text{n}} + 2} \right) \\
  &{{\text{a}}_{\text{n}}} = 3 + {{\text{n}}^2} + {\text{n}} - 2 \\
  &{{\text{a}}_{\text{n}}} = {{\text{n}}^2} + {\text{n}} + 1 \\
\end{align} $
To find the sum of the series, we will add the general terms using summation and get the answer as-
$\mathop \sum \limits_{{\text{r}} = 1}^{\text{n}} {{\text{a}}_{\text{r}}} = \mathop \sum \limits_{{\text{r}} = 1}^{\text{n}} \left( {{{\text{r}}^2} + {\text{r}} + 1} \right)$
We know that-
$\mathop \sum \limits_{{\text{r}} = 1}^{\text{n}} {{\text{r}}^2} = \dfrac{{{\text{n}}\left( {{\text{n}} + 1} \right)\left( {2{\text{n}} + 1} \right)}}{6}$
$\mathop \sum \limits_{{\text{r}} = 1}^{\text{n}} {\text{r}} = \dfrac{{{\text{n}}\left( {{\text{n}} + 1} \right)}}{2}$
$\mathop \sum \limits_{{\text{r}} = 1}^{\text{n}} \left( 1 \right) = {\text{n}}$
Using these properties we can write that-
$\mathop \sum \limits_{{\text{r}} = 1}^{\text{n}} \left( {{{\text{r}}^2} + {\text{r}} + 1} \right) = \mathop \sum \limits_{{\text{r}} = 1}^{\text{n}} {{\text{r}}^2} + \mathop \sum \limits_{{\text{r}} = 1}^{\text{n}} {\text{r}} + \mathop \sum \limits_{{\text{r}} = 1}^{\text{n}} 1$
Applying the values we get-
$\begin{align}
   &= \dfrac{{{\text{n}}\left( {{\text{n}} + 1} \right)\left( {2{\text{n}} + 1} \right)}}{6} + \dfrac{{{\text{n}}\left( {{\text{n}} + 1} \right)}}{2} + {\text{n}} \\
  & = \dfrac{{2{{\text{n}}^3} + {{\text{n}}^2} + 2{{\text{n}}^2} + {\text{n}}}}{6} + \dfrac{{{{\text{n}}^2} + {\text{n}}}}{2} + {\text{n}} \\
  & = \dfrac{{2{{\text{n}}^3} + 3{{\text{n}}^2} + {\text{n}} + 3{{\text{n}}^2} + 3{\text{n}} + 6{\text{n}}}}{6} \\
  & = \dfrac{{2{{\text{n}}^3} + 6{{\text{n}}^2} + 10{\text{n}}}}{6} = \dfrac{{\text{n}}}{3}\left( {{{\text{n}}^2} + 3{\text{n}} + 5} \right) \\
\end{align} $
This is the required answer.

Note: This is a complex question which requires multiple concepts such as AP, GP, sequences and summation. We should always keep in mind that if the common difference of the series itself is making an AP, we should always use method difference.