Question

Please tell me how can I simplify this…?
 \[\dfrac{{{3^n} + {3^{n + 1}}}}{{{3^n} + {3^{n - 1}}}}\]

Answer 1

Hint: Here in this question, we have to simplify the given fractional equation in to the simplest form, these can be solve by many methods, the easiest method is by multiplying 3 on both numerator and denominator on simplify by using law of indices we get the required solution in simplest form.

Complete step by step solution:
The simplest form is the smallest possible equivalent fraction of the number. In fraction, Simplest form is to cancel out the numerator and denominator by a common factor, so that the values cannot be reduced further. But the numerator and denominator still remain as the whole number.
Consider the given fractional equation
 \[\dfrac{{{3^n} + {3^{n + 1}}}}{{{3^n} + {3^{n - 1}}}}\] ----(1)
Here, we have to simplify this into the simplest form.
Multiply both numerator and denominator by 3 in equation (1)
 \[ \Rightarrow \,\,\dfrac{{3\left( {{3^n} + {3^{n + 1}}} \right)}}{{3\left( {{3^n} + {3^{n - 1}}} \right)}}\]
In denominator multiply 3 in to the parenthesis, then we have
 \[ \Rightarrow \,\,\dfrac{{3\left( {{3^n} + {3^{n + 1}}} \right)}}{{3 \cdot {3^n} + 3 \cdot {3^{n - 1}}}}\]
Now, by use the law of indices \[{a^m} \cdot {a^n} = {a^{m + n}}\] in denominator
 \[ \Rightarrow \,\,\dfrac{{3\left( {{3^n} + {3^{n + 1}}} \right)}}{{{3^{n + 1}} + {3^{n - 1 + 1}}}}\]
On simplification, we get
 \[ \Rightarrow \,\,\dfrac{{3\left( {{3^n} + {3^{n + 1}}} \right)}}{{{3^{n + 1}} + {3^n}}}\]
Cancel the common terms in both numerator and denominator, then we get
 \[ \Rightarrow \,\,3\]
Hence we have simplified the given inequality.
Hence, the simplest form of the fraction \[\dfrac{{{3^n} + {3^{n + 1}}}}{{{3^n} + {3^{n - 1}}}}\] is 3.
So, the correct answer is “3”.

Note: The question is in the form of exponential form. We should know about the laws of indices, we apply these laws whenever it is necessary. The above steps will tells how the given term is simplified and hence we have obtained the solution for the given question. Here we have used the law of indices that is \[{a^m} \cdot {a^n} = {a^{m + n}}\] .