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Find the sum of series 1, 3, 5, 7, 9,…… up to n terms?

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Hint: We can see that the given series in the problem follows an Arithmetic Progression (AP). We find the ${{n}^{th}}$ term of the series by using the first term and common difference of the given series. Once, we find the ${{n}^{th}}$ terms, we use the sum of n terms of Arithmetic Progression (AP) to get the desired result.

Complete step-by-step solution:
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 ${{n}^{2}}$.
$\therefore$ The sum of series 1, 3, 5, 7, 9,…… up to n terms is ${{n}^{2}}$.

Note: We can also use the sum of ‘n’ natural numbers to find the sum of the given series after calculating the ${{n}^{th}}$ term. Whenever we see a problem following Arithmetic Progression (AP), we make use of ${{n}^{th}}$ term. Similarly, we can expect problems to find the sum of even numbers up to n-terms.

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