376 ON THE PRODUCT OF ANY TWO LAPLACE'S COEFFICIENTS 



Hence 



n = 2{(r + l)(r + 2)(2n-2r-l)(2n-2r-3)(2n-4r-5)P B _ 2r _ 3 }, 



Continuing the same method of reasoning and applying the method 

 of induction to prove the law for successive terms, we get 



x (2n - 2r - 5)(2n - 4r - 7) P n _ 2r _ 4 }, 

 writing down only the (r+l)th term. 



When n is odd, the last term is when n 2r 4 = 1, so that the term 



n3n-ln+l w . OD 



is -- - -- (w + 4)(n + 2)n. 3P,. 



Wlien n is even, the last term is when n 2r 4 = 0, so that the last 

 term is 



Hence we get the (r+l)th term in the expression for 



(m-l)\ D m P n = ${(r+l)(r + 2) ...(r + m-l)(2n-2r-l)(2n-2r-3) ... 



X (2n - 2r - 2m + 3) (2n - 4r - 2m + 1) P n _ m _ 2r }, 



the last value for n m 2r being or 1, according as n m is even or odd. 

 Hence 



Putting 7i+l for n and m + 1 for m we get 

 l 1. 3. 5 ... 2w-2r+ 1 



3. We have seen above that (2n+ l)-D M P n = Z) m+1 P B+1 -Z) n ' +1 P n _ 1 , and also 

 that (2 + 1 ) /t J D'"P n = (n - m + 1 ) Z>"'P n+1 + (n + m) ZJ-P..,. 



Hence substituting the values given from the above series for D m+1 P n+1 

 and Z) m+1 P n _ 1 , and taking the term involving P n _ m _^ in the series for /iZ) m+1 P B , 

 we get 



- 1 ) ! 1.3.5...(2n-2r-l) 



m! (r- 1)! 1 . 3 . 5 ... (2n-2m-2r-l) 

 (2n - 2r + 1) (n - m} 



(2n-2m-2r + l) 



