Engineering Mathematics: Sum of Binomial Product

Friday, March 9, 2012

Sum of Binomial Product

SBP.ID.1:

$\sum _{k=0}^{pn} (-1)^{k}\binom{pn}{k}\binom{2pn}{2k}^{-1} =\frac{(2pn+1)(1+(-1)^{pn})}{2(pn+1)}$

proof

From Reciprocals of Binomial Coefficients

$\binom{2pn}{2k}^{-1} = (2pn+1)\int _{0}^{1} t^{2k}(1-t)^{2pn-2k} dt$

(1)

Therefore,

$\sum _{k=0}^{pn} (-1)^{k}\binom{pn}{k}\binom{2pn}{2k}^{-1} =\sum _{k=0}^{pn} (-1)^{k}\binom{pn}{k}(2pn+1)\int _{0}^{1} t^{2k}(1-t)^{2pn-2k} dt$

(2)

$\sum _{k=0}^{pn} (-1)^{k}\binom{pn}{k}\binom{2pn}{2k}^{-1} =(2pn+1)\int _{0}^{1} (1-t)^{2pn}\sum _{k=0}^{pn} \binom{pn}{k}(\frac{-t^{2}}{(1-t)^{2}})^{k} dt$

(3)

using the binomial concept ,

$\sum _{k=0}^{pn} (-1)^{k}\binom{pn}{k}\binom{2pn}{2k}^{-1} =(2pn+1)\int _{0}^{1} (1-t)^{2pn} (1-\frac{t^{2}}{(1-t)^{2}})^{pn} dt$

(4)

$\sum _{k=0}^{pn} (-1)^{k}\binom{pn}{k}\binom{2pn}{2k}^{-1} =(2pn+1)\int _{0}^{1} (1-2t)^{pn} dt$

(5)

$\sum _{k=0}^{pn} (-1)^{k}\binom{pn}{k}\binom{2pn}{2k}^{-1} =\frac{(2pn+1)(1+(-1)^{pn})}{2(pn+1)}$

(6)

Engineering Mathematics

Friday, March 9, 2012

Sum of Binomial Product

No comments:

Post a Comment