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    Hi I have in my notes this sum to infinity, starting from i=0, can someone explain it to me:

    (4^(2i))/((2i)!) = (e^4 + e^(-4))/2

    I know that the sum to infinity starting from i = 0

    (4^i)/i! is equal to e^-4

    so could someone explain the above sum to me please.

    Thanks.
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    (Original post by TheBhramaBull)
    Hi I have in my notes this sum to infinity, starting from i=0, can someone explain it to me:

    (4^(2i))/((2i)!) = (e^4 + e^(-4))/2

    I know that the sum to infinity starting from i = 0

    (4^i)/i! is equal to e^-4

    so could someone explain the above sum to me please.

    Thanks.
    Think cosine hyperbolic.

    edit: Alternatively, sum together the Taylor series expansions for  e^x and  e^{-x} , and it should be clear that their sum is twice the sum to infinity in question, for an approproiate choice of x.
 
 
 
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