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Question: For what value of n, \(\dfrac{{{a}^{n+1}}+{{b}^{n+1}}}{{{a}^{n}}+{{b}^{n}}}\) is the harmonic mean o...

For what value of n, an+1+bn+1an+bn\dfrac{{{a}^{n+1}}+{{b}^{n+1}}}{{{a}^{n}}+{{b}^{n}}} is the harmonic mean of a and b?

Explanation

Solution

Hint: Think of the basic definition of Harmonic progression, and equate the given expression with 2aba+b\dfrac{2ab}{a+b} to get the answer.

Complete step-by-step answer:
Before starting with the solution to the above question, let us talk about harmonic progression.
In mathematics, harmonic progression is defined as a sequence of numbers, for which the sequence of the reciprocals of its terms gives an arithmetic progression.
We can represent a general harmonic progression as:1a,1a+d,1a+2d............\dfrac{1}{a},\dfrac{1}{a+d},\dfrac{1}{a+2d}............
Where a,a+d,a+2d...............a,a+d,a+2d............... will represent an arithmetic progression.
Now starting with the solution to the given question.
Harmonic mean = an+1+bn+1an+bn\dfrac{{{a}^{n+1}}+{{b}^{n+1}}}{{{a}^{n}}+{{b}^{n}}}
Also, we know that the harmonic mean of two numbers x and y are given by 2xyx+y\dfrac{2xy}{x+y} . Therefore, using this in the above equation, we get
2aba+b=an+1+bn+1an+bn\dfrac{2ab}{a+b}=\dfrac{{{a}^{n+1}}+{{b}^{n+1}}}{{{a}^{n}}+{{b}^{n}}}
⇒2aba+b=an+1(1+bn+1an+1)an(1+bnan)\Rightarrow \dfrac{2ab}{a+b}=\dfrac{{{a}^{n+1}}\left( 1+\dfrac{{{b}^{n+1}}}{{{a}^{n+1}}} \right)}{{{a}^{n}}\left( 1+\dfrac{{{b}^{n}}}{{{a}^{n}}} \right)}
⇒2ba+b=(1+bn+1an+1)(1+bnan)\Rightarrow \dfrac{2b}{a+b}=\dfrac{\left( 1+\dfrac{{{b}^{n+1}}}{{{a}^{n+1}}} \right)}{\left( 1+\dfrac{{{b}^{n}}}{{{a}^{n}}} \right)}
⇒2ba+b=(1+(ba)n+1)(1+(ba)n)\Rightarrow \dfrac{2b}{a+b}=\dfrac{\left( 1+{{\left( \dfrac{b}{a} \right)}^{n+1}} \right)}{\left( 1+{{\left( \dfrac{b}{a} \right)}^{n}} \right)}
Now we will divide the numerator and denominator of the left-hand side by b. On doing so our expression becomes:
2ab+1=(1+bn+1an+1)(1+bnan)\dfrac{2}{\dfrac{a}{b}+1}=\dfrac{\left( 1+\dfrac{{{b}^{n+1}}}{{{a}^{n+1}}} \right)}{\left( 1+\dfrac{{{b}^{n}}}{{{a}^{n}}} \right)}
On letting ba\dfrac{b}{a} to be k, our expression becomes:
21k+1=(1+kn+1)(1+kn)\dfrac{2}{\dfrac{1}{k}+1}=\dfrac{\left( 1+{{k}^{n+1}} \right)}{\left( 1+{{k}^{n}} \right)}
⇒2kk+1=(1+kn+1)(1+kn)\Rightarrow \dfrac{2k}{k+1}=\dfrac{\left( 1+{{k}^{n+1}} \right)}{\left( 1+{{k}^{n}} \right)}
Now for further solving the equation, we will cross-multiply. This will give us:
2k(1+kn)=(1+kn+1)(1+k)2k\left( 1+{{k}^{n}} \right)=\left( 1+{{k}^{n+1}} \right)\left( 1+k \right)
⇒2k+2kn+1=1+k+kn+1+kn+2\Rightarrow 2k+2{{k}^{n+1}}=1+k+{{k}^{n+1}}+{{k}^{n+2}}
⇒kn+2−kn+1−k+1=0\Rightarrow {{k}^{n+2}}-{{k}^{n+1}}-k+1=0
⇒kn+1(k−1)−1(k−1)=0\Rightarrow {{k}^{n+1}}\left( k-1 \right)-1\left( k-1 \right)=0
⇒(kn+1−1)(k−1)=0\Rightarrow \left( {{k}^{n+1}}-1 \right)\left( k-1 \right)=0
Now we know that a is not equal to b, which means k cannot be equal to 1. Therefore, the only possibility is:
kn+1−1=0{{k}^{n+1}}-1=0
⇒kn+1=1\Rightarrow {{k}^{n+1}}=1
Now, as k is not 1 the power of k should be equal to zero for the equation to be true.
∴n+1=0\therefore n+1=0
⇒n=−1\Rightarrow n=-1
Therefore, the answer to the above question is n = -1.

Note: The key to the above question is letting ba\dfrac{b}{a} to be k. Also, to solve problems like the above one, we need to know the properties and formulas related to different sequences, especially arithmetic and geometric sequence.