Question 42050
{{{e^(ax) = 1 + ax/1! + (ax)^2/2! + (ax)^3/3! + ....}}}{{{+ (ax)^n/n! + .....}}}{{{to}}}{{{infinity}}}

Similarly,
{{{e^(bx) = 1 + bx/1! + (bx)^2/2! + (bx)^3/3! + ....}}}{{{+ (bx)^n/n! + .....}}}{{{to}}}{{{infinity}}}


Multiplying (1) and (2) we have
{{{e^(ax)*e^(bx) }}} = ({{{1 + ax/1! + (ax)^2/2! + (ax)^3/3! + ....}}}{{{+ (ax)^n/n! + .....}}}{{{to}}}{{{infinity}}}) ({{{1 + bx/1! + (bx)^2/2! + (bx)^3/3! + ....}}}{{{+ (bx)^n/n! + .....}}}{{{to}}}{{{infinity}}})


or {{{e^((a+b)x) }}} = ({{{1 + ax/1! + (ax)^2/2! + (ax)^3/3! + ....}}}{{{+ (ax)^n/n! + .....}}}{{{to}}}{{{infinity}}}) ({{{1 + bx/1! + (bx)^2/2! + (bx)^3/3! + ....}}}{{{+ (bx)^n/n! + .....}}}{{{to}}}{{{infinity}}})


Now, let us find the coefficient of {{{x^(2n)}}} in left side.
This is 
[{{{(b^(2n)/(2n)!) + (a/1!)*(b^(2n-1)/(2n-1)!) + ..... }}}{{{+ (a^n/n!)(b^n/n!) +.....}}}{{{ + (a^(2n)/(2n)!)}}}]


Again, coefficient of {{{x^(2n)}}} in the expansion of {{{e^((a+b)x)}}} is {{{(a+b)^2n/(2n)!}}}.

Therefore we can write

{{{(a+b)^(2n)/(2n)! }}} = {{{(b^(2n)/(2n)!) + (a/1!)*(b^(2n-1)/(2n-1)!) + ..... }}}{{{+ (a^n/n!)(b^n/n!) +.....}}}{{{ + (a^(2n)/(2n)!)}}}



Perhaps this is the expression you want.