A) 1
B) \[-1\]
C) 0
D) 2
Correct Answer: B
Solution :
We have \[\frac{{{a}^{n+1}}+{{b}^{n+1}}}{{{a}^{n}}+{{b}^{n}}}=\frac{2ab}{a+b}\] \[\Rightarrow \]\[{{a}^{n+2}}+a{{b}^{n+1}}+b{{a}^{n+1}}+{{b}^{n+2}}=2{{a}^{n+1}}b+2{{b}^{n+1}}a\] \[\Rightarrow \]\[{{a}^{n+1}}(a-b)={{b}^{n+1}}(a-b)\] or \[{{\left( \frac{a}{b} \right)}^{n+1}}=(1)={{\left( \frac{a}{b} \right)}^{0}}\] Hence\[n=-1\].
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