78 Dr. I. J. Schwatt on Methods for the 



Letting r=# 2 , and integrating by parts, we have 

 1 1 — r Tl 1— r, 1 + rh 



Therefore 



(iv.) A fourth method : 



If in the given summation we let r = x 2 , then 



S= n?o(2^ + l)(2n+2)(2n+3)* 

 We then have 



Therefore 



and 



Hence 



= ^[ a?lo sr^ +1 °g( 1 -^]- 

 s = £?£ [* lo «i^ + lo « ^ -^ ] <**• 



Integrating by parts, 



1 py 2 1 H- a? C x x 2 dx , ... .,,1 



1 rl + a; 2 1 + oj i /i ox 1 



= 2s 5 L~2~ g r^ + * g ( ) ~ x \ ' 



(v.) We might also arrive at the result in the following 

 way : 



Let r=# 2 , 



then * ^2«+3 



n r (2n + l)(2n + 2)(2n + 3) 



