InterviewSolution
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InterviewSolution
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Find the value of `n`so that `(a^(n+1)+b^(n+1))/(a^n+b^n)`may be the geometric mean between `aa n dbdot` |
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Answer» ` therefore (a^(n+1)+b^(n+1))/(a^(n)+b^(n))=sqrt(ab)` ` implies (b^(n+1)[((a)/(b))^(n+1)+1])/(b^(n)[((a)/(b))^(n)+1])=bsqrt((a)/(b)) implies (((a)/(b))^(n+1)+1)/(((a)/(b))^(n)+1)=((a)/(b))^((1)/(2))` Let `(a)/(b)= lambda` ` implies (lambda^(n+1)+1)/(lambda^(n)+1)=lambda^((1)/(2)) implies lambda^(n+1)+1=lambda^(n+(1)/(2))+lambda^((1)/(2)` ` implies lambda^(n+(1)/(2))(lambda^((1)/(2))-1)-(lambda^((1)/(2))-1)=0` ` implies (lambda^((1)/(2))-1)(lambda^(n+(1)/(2))-1)=0` ` implies lambda^((1)/(2))-1 ne 0 " " [ therefore a ne b]` ` therefore lambda^(n+(1)/(2))-1=0` `implies lambda^(n+(1)/(2))=1=lambda^(0)` `implies n+(1)/(2)=0 " or "n=-(1)/(2)` |
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