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How i calculate the value of the sum #2^(n+3)/(n!)# ?

1 Answer

Jun 18, 2016

# 2^3e^2#

Explanation:

#e^x = sum_{i=0}^{oo}x^i/(i!) #

so

#e^2 = sum_{i=0}^{oo}2^i/(i!) #

then

#2^3sum_{i=0}^{oo}2^i/(i!) = sum_{i=0}^{oo}2^{i+3}/(i!) = 2^3e^2#

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