Evaluate the following expression \(\sum\limits_{k = 5}^x {\left( {8k + 2} \right)} = 1032\)

The nth  term   of an arithmetic sequence:

\({a_{n{\rm{ }}}} = {\rm{ }}{a_{1{\rm{ }}}} + d\left( {n - 1} \right)\)

Here, in the above expression, the first term is \({a_n}\) , the common difference is d and n is any natural number.

           

A series is formed when the terms of a sequence are added. An arithmetic series is the sum of the terms of an arithmetic sequence. The sum of the first n terms is called the partial sum and is denoted by \({S_n}\). It can be calculated by using following formulas:

              \({S_n} = \frac{n}{2}\left( {{a_1} + {a_n}} \right)\)

Or,
            \({S_n} = \frac{n}{2}\left( {2{a_1} + \left( {n - 1} \right)d} \right)\)

Now consider the given expression,

The sum of series (8k+2) from 5 to x can be expressed as a difference of series (8k+2) from 1 to x and (8k+2) from 1 to 4 as shown below:

\(\begin{array}{c}\sum\limits_{k = 5}^x {\left( {8k + 2} \right)}  = \sum\limits_{k = 1}^x {\left( {8k + 2} \right)}  - \sum\limits_{k = 1}^4 {\left( {8k + 2} \right)} \\1032 = \sum\limits_{k = 1}^x {\left( {8k + 2} \right)}  - \left( {10 + 18 + 26 + 34} \right)\\1032 = \sum\limits_{k = 1}^x {\left( {8k + 2} \right)}  - 88\\1032 + 88 = \sum\limits_{k = 1}^x {\left( {8k + 2} \right)} \end{array}\)

Then, solve for the these,
\(\begin{array}{c}{a_1} = 8\left( 1 \right) + 2\\ = 8 + 2\\ = 10\end{array}\)
\(\begin{array}{c}{a_2} = 8\left( 2 \right) + 2\\ = 16 + 2\\ = 18\end{array}\)
\({a_n} = 8x + 2\)
\(n = x\)
so that,
\(\begin{array}{c}d = {a_2} - {a_1}\\ = 18 - 10\\ = 8\end{array}\)

Use the sum of nterms expression following the steps given below:

\(\begin{array}{c}1120 = \frac{x}{2}(10 + 8x + 2)\\1120 = \frac{x}{2}\left( {8x + 12} \right)\\1120 = \frac{{2x}}{2}\left( {4x + 6} \right)\\1120 = x\left( {4x + 6} \right)\end{array}\)

Further, solve as follow:

\(\begin{array}{c}1120 = 4{x^2} + 6x\\4{x^2} + 6x - 1120 = 0\\4{x^2} - 64x + 70x - 936 = 0\\4x\left( {x - 16} \right) + 70\left( {x - 16} \right) = 0\end{array}\)

\(\begin{array}{c}\left( {x - 16} \right)\left( {4x + 70} \right) = 0\\x - 16 = 0{\rm{ or }}4x + 70 = 0\\x = 16,\frac{{ - 70}}{4}\end{array}\)

Discard the negative value,

x= 16 is the best value for \(\sum\limits_{k = 5}^x {\left( {8k + 2} \right)}  = 1032\).