Infinite springs with force constant k, 2k, 4k, 8k,.. respectively are connected in series. The effective force constant of the spring will be-
$\left( {\text{A}} \right){\text{ }}\dfrac{k}{2}$
$\left( {\text{B}} \right){\text{ }}2k$
$\left( {\text{C}} \right){\text{ }}k$
$\left( {\text{D}} \right){\text{ }}2048k$

Hint:Here we have to find the effective force constant of the spring.
So we use Hooke’s law concept and the effective force constant and then we will find the series by using the series formula.
Finally we get the required answer.

Formulae used:
$F = - kx$ Here F is force; k is spring constant an x is elongation.
${s_\infty } = \dfrac{a}{{1 - r}}$

Complete step by step solution:
Let us consider a 2 spring system, first with spring constant \[{k_1}\] and elongation \[{x_1}\] and second with spring constant \[{k_2}\] an elongation\[{x_2}\].
For both the forces will be the same.
According to Hooke’s law,
$F = - {k_1}{x_1}$
$F = - {k_2}{x_2}$
From Hooke’s law equations, we can write it as,
${x_1} = \dfrac{{ - F}}{{{k_1}}}$ And ${x_2} = \dfrac{{ - F}}{{{k_2}}}$
The total elongation will be
$x = {x_1} + {x_2}$
Substituting the values of \[{x_1}\] and \[{x_2}\] we get
\[x = - F\left( {\dfrac{1}{{{k_1}}} + \dfrac{1}{{{k_2}}}} \right)\]
$\Rightarrow \dfrac{1}{{{k_{eff}}}} = \dfrac{1}{{{k_{_1}}}} + \dfrac{1}{{{k_2}}}$
This formula is applicable for the infinite spring system also
So we can write it as,
$\Rightarrow \dfrac{1}{{{k_{eff}}}} = \dfrac{1}{k} + \dfrac{1}{{2k}} + \dfrac{1}{{4k}} + \dfrac{1}{{8k}}.........$
The above series is in a geometric progression with the first element $a = \dfrac{1}{k}$ and
 $r = \dfrac{1}{{2k}}\dfrac{k}{1}$
On cancelling the term we get
$ \Rightarrow \dfrac{1}{2}$
Here we have to use the formula for a geometric progression, ${S_\infty } = \dfrac{a}{{1 - r}}$
Putting the value in the formula and we get
$\Rightarrow \dfrac{1}{{{k_\infty }}} = \dfrac{{\dfrac{1}{k}}}{{1 - \dfrac{1}{2}}}$
We have to split the term for our convenience and we get,
$\Rightarrow \dfrac{1}{{{k_\infty }}} = \dfrac{1}{k} \times \dfrac{1}{{\dfrac{{2 - 1}}{2}}}$
On subtracting the denominator term we get,
$\Rightarrow \dfrac{1}{{{k_\infty }}} = \dfrac{1}{k} \times \dfrac{1}{{\dfrac{1}{2}}}$
On some simplification we get,
$\dfrac{1}{{{k_\infty }}} = \dfrac{2}{k}$
Hence we can write it as,
${k_\infty } = \dfrac{k}{2}$
Hence the correct option is \[\left( A \right)\] .

Additional information: Hooke’s law is used extensively in the fields of science and technology; it is the basis of seismology, molecular mechanics and acoustics. It is the basic principle behind instruments like spring scale, the manometer and the balanced wheel of a mechanical clock.
The spring was an important invention; it is now used in various machines and instruments. It was invented by Tradwell and has a British patent. There are nearly 10+ types of springs that are used today. The most common are- compression springs, extension springs, torsion spring, constant force spring, Belleville spring, drawbar spring, volute spring and garter spring. The spring is also used in tic-tac pens.

Note:This formula is only applicable when the springs are connected in a series. The formula changes when the springs are connected in parallel. For parallel connection, the effective force constant of the springs is given by-
${k_{eff}} = {k_1} + {k_2}$