Given 2m2b2m b2m2 1m14m22mm2 m23 with b2 14 Find the gener

Given

(2m)2b2m +b2m2 = (1)m+14m/22m(m!)2

m=2,3,...

with b2 = 1/4. Find the general formula for b2m.

Solution

Gven that

(2m)² b_2m + b_2m-2 = [ (-1)^m+1 4m] /[ 2^2m (m!)²]

  (2m)² b_2m = {[ (-1)^m+1 4m] / [2^2m (m!)²]} - b_2m-2

b_2m = {[ (-1)^m+1 4m] / 2^2m (m!)²] - [b_2m-2 ]} / (2m)²

= {[ (-1)^m+1 4m] / 2^2m (m!)²] - [b_2m-2 ]} / 4m²

    =  {[ (-1)^m+1 ] / 2^2m m(m!)²] - [ b_2m-2 / 4m² ]

Therefore,

general formula for b_2m =  {[ (-1)^m+1 ] / 2^2m m(m!)²] - [ b_2m-2 / 4m² ]