
The sum of the series $1+\left( 1+2 \right)+\left( 1+2+3 \right)+.........$ up to $n$ terms will be:
(1) ${{n}^{2}}-2n+6$
(2) $\dfrac{n\left( n+1 \right)\left( 2n-1 \right)}{6}$
(3) ${{n}^{2}}+2n+6$
(4) $\dfrac{n\left( n+1 \right)\left( n+2 \right)}{6}$
Answer
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Hint: Here in this question we have been asked to find the sum of the series $1+\left( 1+2 \right)+\left( 1+2+3 \right)+.........$ up to $n$ terms. We know that the sum of the first $n$ positive natural numbers will be given as $\dfrac{n\left( n+1 \right)}{2}$ . This will be the general term here.
Complete step by step solution:
Now considering from the question we have been asked to find the sum of the series $1+\left( 1+2 \right)+\left( 1+2+3 \right)+.........$ up to $n$ terms.
From the basic concepts, we know that the sum of the first $n$ positive natural numbers will be given as $\dfrac{n\left( n+1 \right)}{2}$ .
Hence we can say that the general term of the series is given as $\dfrac{n\left( n+1 \right)}{2}$ .
Hence the sum of the given series will be given as $\sum\limits_{r=1}^{n}{\dfrac{r\left( r+1 \right)}{2}}$ .
By simplifying this further we will have $\Rightarrow \dfrac{1}{2}\sum\limits_{r=1}^{n}{{{r}^{2}}}+\dfrac{1}{2}\sum\limits_{r=1}^{n}{r}$ .
We know that $\sum\limits_{r=1}^{n}{r}=\dfrac{n\left( n+1 \right)}{2}$ and $\sum\limits_{r=1}^{n}{{{r}^{2}}}=\dfrac{n\left( n+1 \right)\left( 2n+1 \right)}{6}$ . By using these formulae in the expression we have we will get $\Rightarrow \dfrac{1}{2}\left( \dfrac{n\left( n+1 \right)\left( 2n+1 \right)}{6} \right)+\dfrac{1}{2}\left( \dfrac{n\left( n+1 \right)}{2} \right)$ .
By simplifying this further we will have
$\begin{align}
&= \left( \dfrac{n\left( n+1 \right)}{4} \right)\left( \dfrac{2n+1}{3}+1 \right) \\
&= \left( \dfrac{n\left( n+1 \right)}{4} \right)\left( \dfrac{2n+4}{3} \right) \\
&= \dfrac{n\left( n+1 \right)\left( n+2 \right)}{6} \\
\end{align}$ .
Therefore we can conclude that the sum of the series $1+\left( 1+2 \right)+\left( 1+2+3 \right)+.........$ up to $n$ terms will be given as $\dfrac{n\left( n+1 \right)\left( n+2 \right)}{6}$ .
Hence we will mark the option “4” as correct.
Note: In the process of answering questions of this type, we generally get tense by seeing the series directly and lose our confidence and make mistakes due to pressure. This is a very simple question and it can be answered easily. The first thing is to deduct the general term of any given series for answering this type of question.
Complete step by step solution:
Now considering from the question we have been asked to find the sum of the series $1+\left( 1+2 \right)+\left( 1+2+3 \right)+.........$ up to $n$ terms.
From the basic concepts, we know that the sum of the first $n$ positive natural numbers will be given as $\dfrac{n\left( n+1 \right)}{2}$ .
Hence we can say that the general term of the series is given as $\dfrac{n\left( n+1 \right)}{2}$ .
Hence the sum of the given series will be given as $\sum\limits_{r=1}^{n}{\dfrac{r\left( r+1 \right)}{2}}$ .
By simplifying this further we will have $\Rightarrow \dfrac{1}{2}\sum\limits_{r=1}^{n}{{{r}^{2}}}+\dfrac{1}{2}\sum\limits_{r=1}^{n}{r}$ .
We know that $\sum\limits_{r=1}^{n}{r}=\dfrac{n\left( n+1 \right)}{2}$ and $\sum\limits_{r=1}^{n}{{{r}^{2}}}=\dfrac{n\left( n+1 \right)\left( 2n+1 \right)}{6}$ . By using these formulae in the expression we have we will get $\Rightarrow \dfrac{1}{2}\left( \dfrac{n\left( n+1 \right)\left( 2n+1 \right)}{6} \right)+\dfrac{1}{2}\left( \dfrac{n\left( n+1 \right)}{2} \right)$ .
By simplifying this further we will have
$\begin{align}
&= \left( \dfrac{n\left( n+1 \right)}{4} \right)\left( \dfrac{2n+1}{3}+1 \right) \\
&= \left( \dfrac{n\left( n+1 \right)}{4} \right)\left( \dfrac{2n+4}{3} \right) \\
&= \dfrac{n\left( n+1 \right)\left( n+2 \right)}{6} \\
\end{align}$ .
Therefore we can conclude that the sum of the series $1+\left( 1+2 \right)+\left( 1+2+3 \right)+.........$ up to $n$ terms will be given as $\dfrac{n\left( n+1 \right)\left( n+2 \right)}{6}$ .
Hence we will mark the option “4” as correct.
Note: In the process of answering questions of this type, we generally get tense by seeing the series directly and lose our confidence and make mistakes due to pressure. This is a very simple question and it can be answered easily. The first thing is to deduct the general term of any given series for answering this type of question.
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