Finding the Value of n for Harmonic Mean

The question asks us to find the value of \(n\) for which the given algebraic expression represents the harmonic mean of two numbers, \(a\) and \(b\). The given expression is \(\dfrac{a^{n+1} + b^{n+1}}{a^n+b^n}\).

Understanding Harmonic Mean

First, let's recall the formula for the harmonic mean (HM) of two numbers \(a\) and \(b\). The harmonic mean is defined as the reciprocal of the arithmetic mean of the reciprocals of the numbers.

The reciprocals of \(a\) and \(b\) are \(\dfrac{1}{a}\) and \(\dfrac{1}{b}\). The arithmetic mean of these reciprocals is \(\dfrac{\dfrac{1}{a} + \dfrac{1}{b}}{2}\). The reciprocal of this is the harmonic mean.

So, the harmonic mean of \(a\) and \(b\) is:

\[ HM = \dfrac{1}{\dfrac{\dfrac{1}{a} + \dfrac{1}{b}}{2}} = \dfrac{2}{\dfrac{1}{a} + \dfrac{1}{b}} \]

To simplify the denominator:

\[ \dfrac{1}{a} + \dfrac{1}{b} = \dfrac{b}{ab} + \dfrac{a}{ab} = \dfrac{a+b}{ab} \]

So, the harmonic mean is:

\[ HM = \dfrac{2}{\dfrac{a+b}{ab}} = \dfrac{2ab}{a+b} \]

Setting up the Equation to Find the Value of n

We are given that the expression \(\dfrac{a^{n+1} + b^{n+1}}{a^n+b^n}\) is equal to the harmonic mean of \(a\) and \(b\). Therefore, we can set up the equation:

\[ \dfrac{a^{n+1} + b^{n+1}}{a^n+b^n} = \dfrac{2ab}{a+b} \]

We need to find the value of \(n\) that satisfies this equation for any \(a\) and \(b\) (where \(a, b \neq 0\) and \(a+b \neq 0\)).

Testing the Given Options for the Value of n

Let's substitute the possible values of \(n\) from the options into the left side of the equation and see which one makes it equal to the harmonic mean, \(\dfrac{2ab}{a+b}\).

Option 1: \(n = -1\)

Substitute \(n = -1\) into the expression:

\[ \dfrac{a^{-1+1} + b^{-1+1}}{a^{-1}+b^{-1}} = \dfrac{a^0 + b^0}{\dfrac{1}{a}+\dfrac{1}{b}} \]

Since any non-zero number raised to the power of 0 is 1 (\(a^0 = 1, b^0 = 1\)), we get:

\[ \dfrac{1 + 1}{\dfrac{1}{a}+\dfrac{1}{b}} = \dfrac{2}{\dfrac{b+a}{ab}} = \dfrac{2}{\dfrac{a+b}{ab}} \]

Now, invert the denominator and multiply:

\[ 2 \times \dfrac{ab}{a+b} = \dfrac{2ab}{a+b} \]

This matches the harmonic mean formula. So, the value of n = -1 satisfies the condition.

Option 2: \(n = 0\)

Substitute \(n = 0\) into the expression:

\[ \dfrac{a^{0+1} + b^{0+1}}{a^0+b^0} = \dfrac{a^1 + b^1}{a^0+b^0} = \dfrac{a+b}{1+1} = \dfrac{a+b}{2} \]

This is the arithmetic mean of \(a\) and \(b\). This is generally not equal to the harmonic mean \(\dfrac{2ab}{a+b}\) unless \(a=b\).

Option 3: \(n = 1\)

Substitute \(n = 1\) into the expression:

\[ \dfrac{a^{1+1} + b^{1+1}}{a^1+b^1} = \dfrac{a^2 + b^2}{a+b} \]

This expression \(\dfrac{a^2+b^2}{a+b}\) is not equal to the harmonic mean \(\dfrac{2ab}{a+b}\) for arbitrary \(a\) and \(b\).

Option 4: \(n = 1/2\)

Substitute \(n = 1/2\) into the expression:

\[ \dfrac{a^{1/2+1} + b^{1/2+1}}{a^{1/2}+b^{1/2}} = \dfrac{a^{3/2} + b^{3/2}}{a^{1/2}+b^{1/2}} \]

We can factor the numerator using the sum of cubes formula \(x^3+y^3 = (x+y)(x^2-xy+y^2)\), where \(x=a^{1/2}\) and \(y=b^{1/2}\):

\[ a^{3/2} + b^{3/2} = (a^{1/2})^3 + (b^{1/2})^3 = (a^{1/2}+b^{1/2})((a^{1/2})^2 - a^{1/2}b^{1/2} + (b^{1/2})^2) \] \[ = (a^{1/2}+b^{1/2})(a - \sqrt{ab} + b) \]

So the expression becomes:

\[ \dfrac{(a^{1/2}+b^{1/2})(a - \sqrt{ab} + b)}{a^{1/2}+b^{1/2}} = a - \sqrt{ab} + b \]

This expression \(a - \sqrt{ab} + b\) is not equal to the harmonic mean \(\dfrac{2ab}{a+b}\) for arbitrary \(a\) and \(b\).

Conclusion

By testing the options, we found that only the value of n = -1 makes the given expression equal to the harmonic mean of \(a\) and \(b\). This confirms the correct value of n.

The equation \(\dfrac{a^{n+1} + b^{n+1}}{a^n+b^n} = \dfrac{2ab}{a+b}\) is satisfied when the value of n is -1.