Question

Find the value of $1 + \dfrac{1}{4} + \dfrac{{1 \cdot 4}}{{4 \cdot 8}} + \dfrac{{1 \cdot 4 \cdot 7}}{{4 \cdot 8 \cdot 12}} + \ldots \ldots \ldots \infty$.

Answer 1

Here, we need to find the value of the given expression. We will equate the terms of the given expression and the expansion of ${\left( {1 + x} \right)^n}$. We will solve the equations to get the values of $x$ and $n$. Then, we will substitute the values of $x$ and $n$ in the binomial expansion. Further we will simplify the equation to get the required answer.
Formula Used: The binomial expansion of the sum of 1 and a number $x$, raised to the power $n$, is given by the formula ${\left( {1 + x} \right)^n} = 1 + nx + \dfrac{{n\left( {n - 1} \right){x^2}}}{{2!}} + \dfrac{{n\left( {n - 1} \right)\left( {n - 2} \right){x^3}}}{{3!}} + \ldots \infty$.

Complete step by step solution:
We know that the binomial expansion of the sum of 1 and a number $x$, raised to the power $n$, is given by the formula ${\left( {1 + x} \right)^n} = 1 + nx + \dfrac{{n\left( {n - 1} \right){x^2}}}{{2!}} + \dfrac{{n\left( {n - 1} \right)\left( {n - 2} \right){x^3}}}{{3!}} + \ldots \infty {\text{ }} \ldots \ldots \left( 1 \right)$.
Here, $n$ is a real number.
We will equate the terms of the expressions $1 + \dfrac{1}{4} + \dfrac{{1 \cdot 4}}{{4 \cdot 8}} + \dfrac{{1 \cdot 4 \cdot 7}}{{4 \cdot 8 \cdot 12}} + \ldots \ldots \ldots \infty$ and $1 + nx + \dfrac{{n\left( {n - 1} \right){x^2}}}{{2!}} + \ldots \ldots \ldots \infty$.
Comparing the second terms of the expressions, we get
$nx = \dfrac{1}{4}$
Comparing the third terms of the expressions, we get
$\dfrac{{n\left( {n - 1} \right){x^2}}}{{2!}} = \dfrac{{1 \cdot 4}}{{4 \cdot 8}}$
Simplifying the expression, we get
$\Rightarrow \dfrac{{n\left( {n - 1} \right){x^2}}}{2} = \dfrac{1}{8}$
Multiplying both sides of the equation by 8, we get
$\begin{array}{l} \Rightarrow \dfrac{{n\left( {n - 1} \right){x^2}}}{2} \times 8 = \dfrac{1}{8} \times 8\\\ \Rightarrow 4n\left( {n - 1} \right){x^2} = 1\end{array}$
Rewriting the equation, we get
$\Rightarrow 4nx\left( {n - 1} \right)x = 1$
Multiplying $n - 1$ and $x$ in the equation, we get
$\Rightarrow 4nx\left( {nx - x} \right) = 1$
Now, we will substitute the value of $nx$ in the equation.
Substituting $nx = \dfrac{1}{4}$ in the equation, we get
$\Rightarrow 4\left( {\dfrac{1}{4}} \right)\left( {\dfrac{1}{4} - x} \right) = 1$
We will simplify this equation to get the value of $x$.
Thus, we get
$\begin{array}{l} \Rightarrow 1\left( {\dfrac{1}{4} - x} \right) = 1\\\ \Rightarrow \dfrac{1}{4} - x = 1\end{array}$
Rewriting the equation, we get
$\Rightarrow x = \dfrac{1}{4} - 1$
Subtracting the terms of the expression, we get the value of $x$ as
$\Rightarrow x = - \dfrac{3}{4}$
Using the value of $x$, we can find the value of $n$.
Substituting $x = - \dfrac{3}{4}$ in the equation $nx = \dfrac{1}{4}$, we get
$\Rightarrow n\left( {\dfrac{{ - 3}}{4}} \right) = \dfrac{1}{4}$
Multiplying both sides by $- \dfrac{4}{3}$, we get
$\begin{array}{l} \Rightarrow n\left( {\dfrac{{ - 3}}{4}} \right)\left( { - \dfrac{4}{3}} \right) = \dfrac{1}{4}\left( { - \dfrac{4}{3}} \right)\\\ \Rightarrow n = - \dfrac{1}{3}\end{array}$
Therefore, we get the value of $x$ and $n$ as $- \dfrac{3}{4}$ and $- \dfrac{1}{3}$ respectively.
Now, we substitute the value of $x$ and $n$ in equation $\left( 1 \right)$.
Substituting $x = - \dfrac{3}{4}$ and $n = - \dfrac{1}{3}$ in equation $\left( 1 \right)$, we get
${\left[ {1 + \left( { - \dfrac{3}{4}} \right)} \right]^{ - \dfrac{1}{3}}} = 1 + \left( { - \dfrac{1}{3}} \right)\left( { - \dfrac{3}{4}} \right) + \dfrac{{\left( { - \dfrac{1}{3}} \right)\left[ {\left( { - \dfrac{1}{3}} \right) - 1} \right]{{\left( { - \dfrac{3}{4}} \right)}^2}}}{{2!}} + \dfrac{{\left( { - \dfrac{1}{3}} \right)\left[ {\left( { - \dfrac{1}{3}} \right) - 1} \right]\left[ {\left( { - \dfrac{1}{3}} \right) - 2} \right]{{\left( { - \dfrac{3}{4}} \right)}^3}}}{{3!}} + \ldots \infty {\text{ }}$
Simplifying the equation, we get
$\begin{array}{l} \Rightarrow {\left[ {1 - \dfrac{3}{4}} \right]^{ - \dfrac{1}{3}}} = 1 + \dfrac{1}{4} + \dfrac{{\left( { - \dfrac{1}{3}} \right)\left( { - \dfrac{4}{3}} \right)\left( {\dfrac{9}{{16}}} \right)}}{2} + \dfrac{{\left( { - \dfrac{1}{3}} \right)\left( { - \dfrac{4}{3}} \right)\left( { - \dfrac{7}{3}} \right)\left( { - \dfrac{{27}}{{64}}} \right)}}{6} + \ldots \infty {\text{ }}\\\ \Rightarrow {\left[ {\dfrac{1}{4}} \right]^{ - \dfrac{1}{3}}} = 1 + \dfrac{1}{4} + \dfrac{{\dfrac{1}{4}}}{2} + \dfrac{{\dfrac{7}{{16}}}}{6} + \ldots \infty \\\ \Rightarrow {\left[ {\dfrac{1}{4}} \right]^{ - \dfrac{1}{3}}} = 1 + \dfrac{1}{4} + \dfrac{1}{8} + \dfrac{7}{{96}} \ldots \infty \end{array}$
Rewriting $\dfrac{1}{8}$ as $\dfrac{{1 \cdot 4}}{{4 \cdot 8}}$ and $\dfrac{7}{{96}}$ as $\dfrac{{1 \cdot 4 \cdot 7}}{{4 \cdot 8 \cdot 12}}$, we get
$\Rightarrow {\left[ {\dfrac{1}{4}} \right]^{ - \dfrac{1}{3}}} = 1 + \dfrac{1}{4} + \dfrac{{1 \cdot 4}}{{4 \cdot 8}} + \dfrac{{1 \cdot 4 \cdot 7}}{{4 \cdot 8 \cdot 12}} + \ldots \ldots \ldots \infty$
The right side of the equation is the expression we need to find the value of.
Therefore, we have
$\Rightarrow 1 + \dfrac{1}{4} + \dfrac{{1 \cdot 4}}{{4 \cdot 8}} + \dfrac{{1 \cdot 4 \cdot 7}}{{4 \cdot 8 \cdot 12}} + \ldots \ldots \ldots \infty = {\left[ {\dfrac{1}{4}} \right]^{ - \dfrac{1}{3}}}$
Simplifying the expression , we get
$\Rightarrow 1 + \dfrac{1}{4} + \dfrac{{1 \cdot 4}}{{4 \cdot 8}} + \dfrac{{1 \cdot 4 \cdot 7}}{{4 \cdot 8 \cdot 12}} + \ldots \ldots \ldots \infty = {4^{\dfrac{1}{3}}}$

Therefore, the value of the expression $1 + \dfrac{1}{4} + \dfrac{{1 \cdot 4}}{{4 \cdot 8}} + \dfrac{{1 \cdot 4 \cdot 7}}{{4 \cdot 8 \cdot 12}} + \ldots \ldots \ldots \infty$ is ${4^{\dfrac{1}{3}}}$.

Note:
We need to be careful while applying the binomial expansion formula. It is common to make a mistake during the expansion of ${\left( {1 + x} \right)^n}$. Another common mistake we can make is to write the answer as ${\left[ {\dfrac{1}{4}} \right]^{\dfrac{1}{3}}}$. We have to remember that the term is actually ${\left[ {\dfrac{1}{4}} \right]^{ - \dfrac{1}{3}}}$, and not ${\left[ {\dfrac{1}{4}} \right]^{\dfrac{1}{3}}}$. The negative sign in the power allows us to reciprocate the expression to obtain ${4^{\dfrac{1}{3}}}$.