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Hint: The given question is related to arithmetic progression. Try to recall the formulae related to ${{n}^{th}}$ term of an arithmetic progression and the sum of first $n$ terms of an arithmetic progression.

Before proceeding with the solution, we must know the concept of arithmetic progression. Arithmetic progression is a series of numbers in which the difference between any two consecutive numbers is always constant.

Letâ€™s consider an arithmetic progression ${{a}_{1}},{{a}_{2}},{{a}_{3}},{{a}_{4}},{{a}_{5}}.....$ .

We know, the difference between consecutive terms is constant. Let the difference between consecutive numbers be $d$.

So, $d={{a}_{2}}-{{a}_{1}}={{a}_{3}}-{{a}_{2}}={{a}_{4}}-{{a}_{3}}...$

So , we can write ${{a}_{2}}={{a}_{1}}+d$ and ${{a}_{3}}={{a}_{2}}+d={{a}_{1}}+2d$ .

Similarly, ${{a}_{4}}={{a}_{3}}+d={{a}_{1}}+3d$ .

Following the pattern, we can say that the ${{n}^{th}}$ term of the arithmetic progression is given as ${{a}_{n}}={{a}_{1}}+(n-1)d$ .

Now, let the sum of the first $n$ terms of an arithmetic progression be denoted as ${{S}_{n}}$.

So, ${{S}_{n}}={{a}_{1}}+{{a}_{2}}+{{a}_{3}}+....+{{a}_{n}}$.

Substituting the values of ${{a}_{1}},{{a}_{2}},{{a}_{3}}....{{a}_{n}}$ in the expression of ${{S}_{n}}$ , we get

${{S}_{n}}={{a}_{1}}+\left( {{a}_{1}}+d \right)+\left( {{a}_{1}}+2d \right)+\left( {{a}_{1}}+3d \right)+.....+\left( {{a}_{1}}+(n-1)d \right)$

$\Rightarrow {{S}_{n}}=n{{a}_{1}}+d(1+2+3+...+(n-1))$ .

Now, we know, the sum of first $n$ natural numbers is given as $\dfrac{n(n+1)}{2}$ .

So, the sum of first $(n-1)$ natural numbers is $\dfrac{(n-1)n}{2}$

So, ${{S}_{n}}=n{{a}_{1}}+\dfrac{n(n-1)}{2}d$

$\Rightarrow {{S}_{n}}=\dfrac{n}{2}\left[ 2{{a}_{1}}+(n-1)d \right]$ .

Now, we are given an arithmetic progression whose ${{n}^{th}}$ term is given as $(2n+1)$.

To find the first term of the progression, we will substitute $n=1$ in the expression of the ${{n}^{th}}$ term.

So, the first term is given as ${{a}_{1}}=(2\times 1)+1=3$.

Now, to find the common difference, we must know the value of ${{2}^{nd}}$ term of the progression. To find the value of the second term, we will substitute $n=2$ in the expression of the ${{n}^{th}}$ term.

So, the second term is given as ${{a}_{2}}=\left( 2\times 2 \right)+1=5$.

Now, the common difference $d={{a}_{2}}-{{a}_{1}}=5-3=2$.

So, we have an arithmetic progression with the first term ${{a}_{1}}=3$ and common difference $d=2$ .

So, the sum of the first $n$ terms of the given arithmetic progression is given as ${{S}_{n}}=\dfrac{n}{2}\left[ (2\times 3)+(n-1)2 \right]$

$=\dfrac{n}{2}\left( 6+2n-2 \right)$

$=\dfrac{n}{2}(4+2n)$

$=n(n+2)$

Hence, the sum of the first $n$ terms of an arithmetic progression, whose ${{n}^{th}}$ term is given by $(2n+1)$, is given as ${{S}_{n}}=n(n+2)$.

Note: The common difference of an arithmetic progression is given as $d={{a}_{2}}-{{a}_{1}}$ and not $d={{a}_{1}}-{{a}_{2}}$. Students generally get confused and make this mistake.

Before proceeding with the solution, we must know the concept of arithmetic progression. Arithmetic progression is a series of numbers in which the difference between any two consecutive numbers is always constant.

Letâ€™s consider an arithmetic progression ${{a}_{1}},{{a}_{2}},{{a}_{3}},{{a}_{4}},{{a}_{5}}.....$ .

We know, the difference between consecutive terms is constant. Let the difference between consecutive numbers be $d$.

So, $d={{a}_{2}}-{{a}_{1}}={{a}_{3}}-{{a}_{2}}={{a}_{4}}-{{a}_{3}}...$

So , we can write ${{a}_{2}}={{a}_{1}}+d$ and ${{a}_{3}}={{a}_{2}}+d={{a}_{1}}+2d$ .

Similarly, ${{a}_{4}}={{a}_{3}}+d={{a}_{1}}+3d$ .

Following the pattern, we can say that the ${{n}^{th}}$ term of the arithmetic progression is given as ${{a}_{n}}={{a}_{1}}+(n-1)d$ .

Now, let the sum of the first $n$ terms of an arithmetic progression be denoted as ${{S}_{n}}$.

So, ${{S}_{n}}={{a}_{1}}+{{a}_{2}}+{{a}_{3}}+....+{{a}_{n}}$.

Substituting the values of ${{a}_{1}},{{a}_{2}},{{a}_{3}}....{{a}_{n}}$ in the expression of ${{S}_{n}}$ , we get

${{S}_{n}}={{a}_{1}}+\left( {{a}_{1}}+d \right)+\left( {{a}_{1}}+2d \right)+\left( {{a}_{1}}+3d \right)+.....+\left( {{a}_{1}}+(n-1)d \right)$

$\Rightarrow {{S}_{n}}=n{{a}_{1}}+d(1+2+3+...+(n-1))$ .

Now, we know, the sum of first $n$ natural numbers is given as $\dfrac{n(n+1)}{2}$ .

So, the sum of first $(n-1)$ natural numbers is $\dfrac{(n-1)n}{2}$

So, ${{S}_{n}}=n{{a}_{1}}+\dfrac{n(n-1)}{2}d$

$\Rightarrow {{S}_{n}}=\dfrac{n}{2}\left[ 2{{a}_{1}}+(n-1)d \right]$ .

Now, we are given an arithmetic progression whose ${{n}^{th}}$ term is given as $(2n+1)$.

To find the first term of the progression, we will substitute $n=1$ in the expression of the ${{n}^{th}}$ term.

So, the first term is given as ${{a}_{1}}=(2\times 1)+1=3$.

Now, to find the common difference, we must know the value of ${{2}^{nd}}$ term of the progression. To find the value of the second term, we will substitute $n=2$ in the expression of the ${{n}^{th}}$ term.

So, the second term is given as ${{a}_{2}}=\left( 2\times 2 \right)+1=5$.

Now, the common difference $d={{a}_{2}}-{{a}_{1}}=5-3=2$.

So, we have an arithmetic progression with the first term ${{a}_{1}}=3$ and common difference $d=2$ .

So, the sum of the first $n$ terms of the given arithmetic progression is given as ${{S}_{n}}=\dfrac{n}{2}\left[ (2\times 3)+(n-1)2 \right]$

$=\dfrac{n}{2}\left( 6+2n-2 \right)$

$=\dfrac{n}{2}(4+2n)$

$=n(n+2)$

Hence, the sum of the first $n$ terms of an arithmetic progression, whose ${{n}^{th}}$ term is given by $(2n+1)$, is given as ${{S}_{n}}=n(n+2)$.

Note: The common difference of an arithmetic progression is given as $d={{a}_{2}}-{{a}_{1}}$ and not $d={{a}_{1}}-{{a}_{2}}$. Students generally get confused and make this mistake.

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