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According to the problem, we have a series given as 1, 3, 5, 7, 9,……. We need to find the sum of the elements in this series up to n terms.

We know that an Arithmetic Progression (AP) is of form a, a+d, a+2d,…….., where ‘a’ is known as the first term and ‘d’ is known as common difference.

We can see that the difference between any two consecutive numbers is 2. We can see that the given series are in AP (Arithmetic Progression) with the first term ‘1’ and common difference ‘2’.

Let us find the ${{n}^{th}}$ term for the given series. We know that ${{n}^{th}}$ term of an Arithmetic Progression (AP) is defined as ${{T}_{n}}=a+\left( n-1 \right)d$.

So, we have ${{n}^{th}}$ term for the given series as ${{T}_{n}}=1+\left( n-1 \right)2$.

${{T}_{n}}=1+2n-2$.

${{T}_{n}}=2n-1$ ---(1).

Now let us find the sum of n terms of the given series. We know that sum of n terms of an Arithmetic Progression (AP) is defined as ${{S}_{n}}=\dfrac{n}{2}\times \left( a+{{T}_{n}} \right)$.

So, we have sum of n terms of the given series as ${{S}_{n}}=\dfrac{n}{2}\times \left( 1+2n-1 \right)$.

${{S}_{n}}=\dfrac{n}{2}\times \left( 2n \right)$.

${{S}_{n}}=n\times n$.

${{S}_{n}}={{n}^{2}}$.

We have found the sum of n terms of the series 1, 3, 5, 7, …… is ${{n}^{2}}$.

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