Based upon this info, we are to find $a$-$a^2$+$a^3$-$a^4$... I could find that $a$=$b$/$(1-b)$ as it is infinite geometric progression. I tried doing mathematical induction but was stuck with a very tedious answer. Is there any better method for the same? Answer is b.
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Since $a=\frac b{1-b}$,$$a-a^2+a^3-a^4+\cdots=\frac a{1+a}=\frac{\frac b{1-b}}{1+\frac b{1-b}}=b.$$
José Carlos Santos
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I think I got it where I missed the point. In the series a-a^2+a^2...ratio will be -a. Thank you. This solves it. – Minhaz Pathan Apr 07 '18 at 13:54
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@MinhazPathan If my answer was useful, perhaps that you could mark it as the accepted one. – José Carlos Santos Apr 07 '18 at 14:13
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How do I do that? – Minhaz Pathan Apr 07 '18 at 14:31
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I have ticked the swoosh. Is that equivalent to marked as accepted one? – Minhaz Pathan Apr 07 '18 at 14:32
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@MinhazPathan Yes, that's it. – José Carlos Santos Apr 07 '18 at 14:33
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$a-a^2+a^3-a^4+\cdots=a+a(-a)+a(-a)^2+a(-a)^3+\cdots$ is also an infinite geometric series.
Note that $\displaystyle a=\frac{b}{1-b}=\frac{1}{1-b}-1$. When $\displaystyle -\frac{1}{2}<b<\frac{1}{2}$,
$$-\frac{1}{3}=\frac{1}{1-\frac{-1}{2}}-1<\frac{1}{1-b}-1<\frac{1}{1-\frac{1}{2}}-1=1$$
$$a-a^2+a^3-a^4+\cdots=\frac{a}{1-(-a)}=b$$
CY Aries
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