3it has been sometime since this was posted.
i guess this is too tough???
\dpi{120} \hspace{-16}\frac{\binom{2010}{0}}{1.1}+\frac{\binom{2010}{2}}{3.4}+\frac{\binom{2010}{4}}{5.4^2}+\frac{\binom{2010}{6}}{7.4^3}+..................+\frac{\binom{2010}{2010}}{2011.4^{1005}}=\\\\\\ $Where $\binom{n}{r}=\frac{n!}{r!.(n-r)!}$
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2 Answers
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262(1+x)^{2010}= C_{0 }+ C_{1}x +C_{2}x^{2} +....... + C_{2010}x^{2010}
integrating both sides,
\frac{(1+x)^{2011} -1}{2011}= C_{0}x+ C_{1}x^{2}/2 +C_{2}x^{3}/3 +....... + C_{2010}x^{2011}/2011
putting x=1/2 and -1/2 ,
\frac{(1+1/2)^{2011} -1}{2011}= C_{0}(1/2)+ C_{1}(1/2)^{2}/2 +C_{2}(1/2)^{3}/3 +....... + C_{2010}(1/2)^{2011}/2011
(eqn 1)
\frac{(1-1/2)^{2011} -1}{2011}= C_{0}(-1/2)+ C_{1}(-1/2)^{2}/2 +C_{2}(-1/2)^{3}/3 +..... + C_{2010}(-1/2)^{2011}/2011
(eqn2)
now (eqn 1) - (eqn2) ,
\frac{(3/2)^{2011} -(1/2)^{2011}}{2011}= 2[C_{0}(1/2)+C_{2}(1/2)^{3}/3 +....... + C_{2010}(1/2)^{2011}/2011]
\frac{(3/2)^{2011} -(1/2)^{2011}}{2011}= [C_{0}/(1.1)+C_{2}/(3.4) +....... + C_{2010}/(2011.4^{1005})]
