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

Hint:
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 \[{4^{\dfrac{1}{3}}}\].