Question

Find the sum to $n$ terms of the series ${1^2} + {3^2} + {5^2} + ...$ to $n$ terms

Answer 1

Here we are asked to find the sum of the series to $n$ terms ${1^2} + {3^2} + {5^2} + \ldots \;to\;n\;terms$. First, we need to find the ${n^{th}}$ term of the given series. The ${n^{th}}$ term of the given series is the generalized form of the given series. Then while simplifying, we need to apply the required formulae.

Formula to be used:
${\left( {a - b} \right)^2} = {a^2} - 2ab + {b^2}$
The sum of the natural numbers is $\sum {n = \dfrac{{n\left( {n + 1} \right)}}{2}}$
The sum of the squares of the natural numbers is $\sum {{n^2}} = \dfrac{{n \times (n + 1) \times (2n + 1)}}{6}$

Complete step-by-step answer:
The given series is ${1^2} + {3^2} + {5^2} + \ldots \;to\;n\;terms$
Here we are asked to calculate the sum of the given series ${1^2} + {3^2} + {5^2} + \ldots \;to\;n\;terms$
Let us assume that ${a_n}$ be ${n^{th}}$ a term of the given series.
We can note that the series ${1^2} + {3^2} + {5^2} + \ldots \;to\;n\;terms$ is in the form ${a_n} = {\left( {2n - 1} \right)^2}$
We are also able to verify ${a_n} = {\left( {2n - 1} \right)^2}$.
When we substitute $n = 1$ in ${a_n} = {\left( {2n - 1} \right)^2}$, we get the first term ${1^2}$ .
Similarly, when we substitute $n = 2$ in${a_n} = {\left( {2n - 1} \right)^2}$, we get the second term${3^2}$.
Similarly, when we substitute $n = 3$ in ${a_n} = {\left( {2n - 1} \right)^2}$, we get the third term ${5^2}$.
Hence, we verified.
Thus, we have ${a_n} = {\left( {2n - 1} \right)^2}$where ${a_n}$is the ${n^{th}}$ term of the given series.
${a_n} = {\left( {2n - 1} \right)^2}$
$= 4{n^2} - 4n + 1$ (Here we have applied the algebraic identity ${\left( {a - b} \right)^2} = {a^2} - 2ab + {b^2}$)
Thus, ${a_n} = 4{n^2} - 4n + 1$
We usually denote the sum of the series by the summation $\sum$.
Here we shall denote the sum to n terms of the given series as $\sum {{a_n}}$ .
Let us substitute ${a_n} = 4{n^2} - 4n + 1$.
$\sum {{a_n}} = \sum {\left( {4{n^2} - 4n + 1} \right)}$
$= \sum {4{n^2}} - \sum {4n} + \sum 1$ (Here we multiplied throughout by summation.)
$= 4\sum {{n^2}} - 4\sum n + \sum 1$ (Now we have taken the constant term outside)
Usually, the summation of a constant is equal to n times the constant.
Hence, $\sum {1 = n}$ .
$\Rightarrow \sum {{a_n}} = 4\sum {{n^2}} - 4\sum n + n$
$= 4\dfrac{{n \times (n + 1) \times (2n + 1)}}{6} - 4\dfrac{{n\left( {n + 1} \right)}}{2} + n$ (Here we applied the formulae) $= 2\dfrac{{\left( {{n^2} + n} \right)(2n + 1)}}{3} - 2\left( {{n^2} + n} \right) + n$
$= 2\dfrac{{(2{n^3} + {n^2} + 2{n^2} + n)}}{3} - 2{n^2} - 2n + n$
$= 2\dfrac{{(2{n^3} + 3{n^2} + n)}}{3} - 2{n^2} - n$
$= \dfrac{{4{n^3} + 6{n^2} + 2n - 6{n^2} - 3n}}{3}$
$= \dfrac{{4{n^3} - n}}{3}$
$= \dfrac{{n\left( {4{n^2} - 1} \right)}}{3}$
Therefore, $\sum {{a_n}} = \dfrac{{n\left( {4{n^2} - 1} \right)}}{3}$
Hence the sum of the series to n terms is $\dfrac{{n\left( {4{n^2} - 1} \right)}}{3}$ .

Note: Generally, a series refers to the sum of the sequence. Here we can simplify the obtained answers further. We need to apply the algebraic identity ${a^2} - {b^2} = \left( {a - b} \right)\left( {a + b} \right)$.
Thus, we get $\dfrac{{n\left( {4{n^2} - 1} \right)}}{3} = \dfrac{{n\left( {2n - 1} \right)\left( {2n + 1} \right)}}{3}$ (Here $a = 2n$ and $b = 1$ )
Hence the sum of the series to n terms is $\dfrac{{n\left( {2n - 1} \right)\left( {2n + 1} \right)}}{3}$ .