Let's say there's a series of the form $$S=\frac{1}{10^2}+\frac{1\cdot3}{1\cdot2\cdot10^4}+\frac{1\cdot3\cdot5}{1\cdot2\cdot3\cdot10^6}+...$$ Now i had written the rth term as $$T_r=\frac{1\cdot3\cdot5....(2r-1)}{1\cdot2\cdot3.... r\cdot10^{2r}}=\frac{2r!}{r!\cdot r!\cdot2^r\cdot10^{2r}}$$ I came to the second equivalence by mutliplying and dividing the first expression with $2\cdot4\cdot6....2r\;$and then taking out a power of 2 from each of the even numbers multiplied in the denomininator. From the looks of it, these expressions tend to give the idea of being solved using binomial most probably the expansion for negative indices but I don't understand how to get to the result from here
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Is it supposed to be $10^6$ in the last denominator in the expression for $S$? – Arthur Nov 16 '17 at 07:49
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yeah, edited in question – Siddharth Jha Nov 16 '17 at 09:10
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$$(1+x)^{-1/2}=1-\frac x2+\frac{3x^2}{2^22!} -\frac{3\cdot 5x^3}{2^32!}+\frac{3\cdot5\cdot7 x^4}{2^44!}-\cdots.$$ Can you choose a suitable $x$ to match this up with your series?
Angina Seng
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