Question

The value of the expression $\dfrac{{1 + 4.343 + 7.4 + 2.3.49 + 7.343}}{{16 + {2^6}{{.3}^1} + {2^5}{{.3}^3} + {2^6}{{.3}^3} + {2^4}{{.3}^4}}}$ equal
A.1
B.2
C.4
D.3

Answer 1

Hint : In this question, we have been given an expression whose value is to be determined. Looking at the expression at first, it seems impossible to solve without actually solving each expression completely. But that is not the case. We always have to remember that if an expression is very lengthy, then there is always a way to simplify it and then solve it. We just have to figure out how. One thing to note in the expression is that in both of the numerator and the denominator, the first term and the last term is a number of type ${x^4}$ , i.e., it is a fourth power of a number. Also, in this question’s expression, it is required that we have to rearrange the terms in both of the numerator and the denominator, and the shifting of powers is also required.
Formula Used:
In this question, we are going to use the binomial theorem, which is:
${\left( {x + y} \right)^n} = \sum\limits_{k = 0}^n {\left( \begin{array}{l}n\\\k\end{array} \right){x^{n - k}}{y^k}} = \sum\limits_{k = 0}^n {\left( \begin{array}{l}n\\\k\end{array} \right){x^k}{y^{n - k}}}$

Complete step-by-step answer :
Now, the given expression $x = \dfrac{{1 + 4.343 + 7.4 + 2.3.49 + 7.343}}{{16 + {2^6}{{.3}^1} + {2^5}{{.3}^3} + {2^6}{{.3}^3} + {2^4}{{.3}^4}}}$ can be rearranged as:
$x = \dfrac{{1 + 7.4 + 2.3.49 + 4.343 + 7.343}}{{16 + {2^6}{{.3}^1} + {2^5}{{.3}^3} + {2^6}{{.3}^3} + {2^4}{{.3}^4}}}$
Now, by shifting the powers (for e.g., ${2^3}{.3^5}$ can be written as ${6^3}{.3^2}$ ), we have:
$x = \dfrac{{{{1.1}^4}{{.7}^0} + {{4.1}^3}{{.7}^1} + {{6.1}^2}{{.7}^2} + {{4.1}^1}{{.7}^3} + {{1.1}^0}{{.7}^4}}}{{{{1.2}^4}{{.6}^0} + {{4.2}^3}{{.6}^1} + {{6.2}^2}{{.6}^2} + {{4.2}^1}{{.6}^3} + {{1.2}^0}{{.6}^4}}}$
Now, clearly the above expression makes more sense and can be written as:
$x = \dfrac{{^4{C_0}{{.1}^4}{{.7}^0}{ + ^4}{C_1}{{.1}^3}{{.7}^1}{ + ^4}{C_2}{{.1}^2}{{.7}^2}{ + ^4}{C_3}{{.1}^1}{{.7}^3}{ + ^4}{C_4}{{.1}^0}{{.7}^4}}}{{^4{C_0}{{.2}^4}{{.6}^0}{ + ^4}{C_1}{{.2}^3}{{.6}^1}{ + ^4}{C_2}{{.2}^2}{{.6}^2}{ + ^4}{C_3}{{.2}^1}{{.6}^3}{ + ^4}{C_4}{{.2}^0}{{.6}^4}}}$
w.k.t
${\left( {x + y} \right)^n} = \sum\limits_{k = 0}^n {\left( \begin{array}{l}n\\\k\end{array} \right){x^{n - k}}{y^k}} = \sum\limits_{k = 0}^n {\left( \begin{array}{l}n\\\k\end{array} \right){x^k}{y^{n - k}}}$

Comparing the two equations and now, this expression can be simplified into:
$x = \dfrac{{{{\left( {1 + 7} \right)}^4}}}{{{{\left( {2 + 6} \right)}^4}}} = \dfrac{{{8^4}}}{{{8^4}}} = 1$
Hence, the answer of the given expression is $1$ .
So, the correct answer is “Option A”.

Note : So, we saw that in solving questions like these, it is always a nice practice if we first write what has been given to us. If the expression is too lengthy or complicated, we must always remember that there is always a way to solve it. We just have to figure out how. Other things like rearranging the terms, shifting the powers, et cetera are also required to be done in the question, depending on the type of the question.