a-n-1-b-n-1-a-n-b-n-ab-find-n-

Question Number 178240 by Shrinava last updated on 14/Oct/22

\frac{a^{n + 1} + b^{n + 1}}{a^{n} + b^{n}} = \sqrt{ab} find n = ?

Answered by Rasheed.Sindhi last updated on 14/Oct/22

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

a^{n + 1} + b^{n + 1} = a^{n + \frac{1}{2}} b^{\frac{1}{2}} + a^{\frac{1}{2}} b^{n + \frac{1}{2}}

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

a^{n + \frac{1}{2}} \overset{\times}{(a^{\frac{1}{2}} - b^{\frac{1}{2}})} = b^{n + \frac{1}{2}} \overset{\times}{(a^{\frac{1}{2}} - b^{\frac{1}{2}})} [a \neq b]

a^{n + \frac{1}{2}} = b^{n + \frac{1}{2}}

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

n + \frac{1}{2} = 0

n = - \frac{1}{2}

Commented by mr W last updated on 14/Oct/22

n i c e s o l u t i o n!

Commented by Rasheed.Sindhi last updated on 14/Oct/22

T h a n k s s i r!

Commented by Tawa11 last updated on 14/Oct/22

Great sir

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