Find the value of n so that \( \dfrac{a^{n+1}+b^{n+1}}{a^n+b^n}\)  may be the geometric mean between a and b.

Asked by Abhisek | 1 year ago |  68

1 Answer

Solution :-

The geometric mean of a and b is \( \sqrt{ab}\) .

According to conditions given in the question,

\( \dfrac{a^{n+1}+b^{n+1}}{a^n+b^n}\)

Square on both the sides.

\( \dfrac{(a^{n+1}+b^{n+1})^2}{(a^n+b^n)^2}=ab\)

\( a^{2n+2}+2a^{n+1}b^{n+1}+b^{2n+2}\)

\( (ab)(a^{2n}+2a^nb^n+b^{2n})\)

\( a^{2n+2}+2a^{n+1}b^{n+1}+b^{2n+2}\)

\( a^{2n+2}+2a^{n+1}b^{n+1}+ab^{2n+1}\)

\( a^{2n+2}+b^{2n+2}=a^{2n+1}+ab^{2n+1}\)

\( a^{2n+2}-b^{2n+1}b=ab^{2n+1}-b^{2n+2}\)

\( a^{2n+1}-=(a-b)=b^{2n+1}(a-b)\)

\(( \dfrac{a}{b})^{2n+1}=1=(\dfrac{a}{b})^0\)

\( 2n+1=0\)

\( n=\dfrac{-1}{2}\)

Answered by Abhisek | 1 year ago

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