Question

If (a^n-1 + b^n-1)/ a^n + b^n is the arithmetic mean between a & b...
Find n..

Answer 1

$\sf \implies 2(a^n \: + \: b^n ) \: = \: (a \: + \: b)(a^{n \: - \: 1} \: + \: b^{n \: - \: 1 }) \\ \\ \\ \sf \implies 2a^n \: + \: 2b^n \: = \: a^n \: + \: ab^{(n \: - \: 1 )} \: + \: ba^{(n \: - \: 1 )} \: + \: b^n \\ \\ \\ \sf \implies 2a^n \: - \: a^n \: + \: 2b^n \: - \: b^n \: = \: ab^{(n \: - \: 1 )} \: + \: ba^{(n \: - \: 1 )} \\ \\ \\ \sf \implies {a}^{n} \: + \: {b}^{n} \: = \: a {b}^{(n \: - \: 1)} \: + \: b {a}^{(n \: - \: 1)} \\ \\ \\ \sf \implies {a}^{n} \: - \: b {a}^{(n \: - \: 1)} \: = \: a {b}^{(n \: - \: 1)} \: - \: {b}^{n}$

$\sf \implies {a}^{(n \: - 1)} \cancel{ \{a \: - \: b \}}\: = \: b^{(n \: - \: 1)} \cancel{ \{ a \: - \: b\}} \\ \\ \\ \sf \implies {a}^{(n \: - \: 1)} \: = \: {b}^{(n \: - \: 1) } \\ \\ \\ \sf \implies \dfrac{ {a}^{(n \: - \: 1)} }{ {b}^{(n \: - \: 1)} } \: = \: 1 \\ \\ \\ \sf \implies \left \{\dfrac{a}{b} \right \}^{(n \: - \: 1)} \: = \: 1$

$\sf \implies \left \{\dfrac{a}{b} \right \}^{(n \: - \: 1)} \: = \: \left \{\dfrac{a}{b} \right \}^{0}$

$\underline{\textsf{On Comparison,}} \\ \\ \sf \implies n \: - \: 1 \: = \: 0 \\ \\ \\ \sf \: \: \: \therefore \: \: n \: = \: 1$