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If `A=1+r^a+r^(2a)+`to `ooa n dB=1+r^b+r^(2b)+oo`, prove that`r=((A-1)/A)^(1//a)=((B-1)/B)^(1//a)` |
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Answer» By summing the given infinite geometric series, we get `x=1/((1-r^(a))) and y=1/((1-r^(b)))` `rArr (1-r^(a))=1/x` and `(1-r^(b))=1/y` `rArr r^(a)=(1-1/x) and r^(b)=(1-1/y)` `rArr r=((x-1)/x)^(1/a) and r=((y-1)/y)^(1/b)`. Hence, `r=((x-1)/x)^(1/a)=((y-1)/y)^(1/b)`. |
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