$\displaystyle \frac{4}{20}$+$\displaystyle \frac{4.7}{20.30}$+$\displaystyle \frac{4.7.10}{20.30.40}$+...
Now I have tried to solve this in a usual way, first find the nth term $t_n$.
$t_n$= $\displaystyle \frac{1}{10}$($\displaystyle \frac{1+3}{2}$) + $\displaystyle \frac{1}{10^2}$($\displaystyle \frac{1+3}{2}$)($\displaystyle \frac{1+6}{3}$) + ...+ $\displaystyle \frac{1}{10^n}$($\displaystyle \frac{1+3}{2}$)($\displaystyle \frac{1+6}{3}$)...($\displaystyle \frac{1+3n}{n+1}$)
=$\displaystyle \frac{1}{10^n}\prod$(1+$\displaystyle \frac{2r}{r+1}$) , $r=1,2,..,n$
=$\displaystyle \prod$($\displaystyle \frac{3}{10}-\displaystyle \frac{1}{5(r+1)}$)
thus, $t_n=$ (x-$\displaystyle \frac{a}{2}$)(x-$\displaystyle \frac{a}{3}$)...(x-$\displaystyle \frac{a}{n+1}$), x=$\displaystyle \frac{3}{10}$, a=$\displaystyle \frac{1}{5}$
Now to calculate $S_n$, I have to find the product $t_n$, and then take sum over it. But this seems to be a very tedious job. Is there any elegant method(may be using the expansions of any analytic functions) to do this?
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