Question

$1 + 3 + 6 + 10 + .....$ up to $n$ terms is equal to
A) $\dfrac{1}{3}n\left( {n + 1} \right)\left( {n + 2} \right)$
B) $\dfrac{1}{6}n\left( {n + 1} \right)\left( {n + 2} \right)$
C) $\dfrac{1}{{12}}n\left( {n + 2} \right)\left( {n + 3} \right)$
D) $\dfrac{1}{{12}}n\left( {n + 1} \right)\left( {n + 2} \right)$

Answer 1

Here we have to find the sum of the given series. For that, we will first convert the terms in general form i.e. in the form of the variable. Then we will find the sum of the general term of this series using sigma. We will use the formula of the sum of special series to find the resultant sum.
Formula used:
We will use the following formulas:

1. $\sum\limits_{k = 1}^n {{k^2}} = \dfrac{{n\left( {n + 1} \right)\left( {2n + 1} \right)}}{6}$
2. $\sum\limits_{k = 1}^n {k = \dfrac{{n\left( {n + 1} \right)}}{2}}$

Complete Step by Step Solution:
Here we have to find the sum of the given series i.e. the sum of $1 + 3 + 6 + 10 + .....$ up to $n$ terms.
Let ${S_n}$ be the sum of the given series.
We can write the last term of series as
$\Rightarrow {S_n} = 1 + 3 + 6 + 10 + .....\dfrac{{n\left( {n + 1} \right)}}{2}$
The terms of this series is of the form $\dfrac{{k\left( {k + 1} \right)}}{2}$.
If we find the sum of term $\dfrac{{k\left( {k + 1} \right)}}{2}$ using sigma where $k$ varies from 1 to $n$, we will get the resultant sum of the series.
Therefore,
$\Rightarrow {S_n} = \sum\limits_{k = 1}^n {\dfrac{{k\left( {k + 1} \right)}}{2}}$
On multiplying the terms, we get
$\Rightarrow {S_n} = \sum\limits_{k = 1}^n {\dfrac{{{k^2} + k}}{2}}$
On further simplification, we get
$\Rightarrow {S_n} = \dfrac{1}{2}\left( {\sum\limits_{k = 1}^n {{k^2}} + \sum\limits_{k = 1}^n k } \right)$ …….. $\left( 1 \right)$
We know the formula of sum of series $\sum\limits_{k = 1}^n {{k^2}} = \dfrac{{n\left( {n + 1} \right)\left( {2n + 1} \right)}}{6}$ and $\sum\limits_{k = 1}^n {k = \dfrac{{n\left( {n + 1} \right)}}{2}}$ .
We will substitute the value of sum of both the series in equation $\left( 1 \right)$. Therefore, we get
$\Rightarrow {S_n} = \dfrac{1}{2}\left( {\dfrac{{n\left( {n + 1} \right)\left( {2n + 1} \right)}}{6} + \dfrac{{n\left( {n + 1} \right)}}{2}} \right)$
On taking common terms out of the bracket, we get
$\Rightarrow {S_n} = \dfrac{1}{2} \times \dfrac{{n\left( {n + 1} \right)}}{2}\left( {\dfrac{{\left( {2n + 1} \right)}}{3} + 1} \right)$
On adding the terms inside the bracket, w get
$\begin{array}{l} \Rightarrow {S_n} = \dfrac{1}{2} \times \dfrac{{n\left( {n + 1} \right)}}{2}\left( {\dfrac{{2n + 1 + 3}}{3}} \right)\\\ \Rightarrow {S_n} = \dfrac{1}{2} \times \dfrac{{n\left( {n + 1} \right)}}{2}\left( {\dfrac{{2n + 4}}{3}} \right)\end{array}$
On further simplification, we get
$\Rightarrow {S_n} = \dfrac{{n\left( {n + 1} \right)\left( {n + 2} \right)}}{6}$
Hence, the sum of the given series is equal to $\dfrac{{n\left( {n + 1} \right)\left( {n + 2} \right)}}{6}$.

Thus, the correct option is option B.

Note:
Here to find the sum of such series, we need to remember the formula of sum of some special series. Special series are defined as the series whose sum can be expressed by explicit formulae. Common examples of special series are the sum of squares of the natural numbers, sum of cubes of the natural numbers etc. Here we might make a mistake by thinking of the series in AP and then using the formula of sum of $n$ terms of AP. This will be wrong because it is nowhere given that the series is in AP.