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2 years ago

∑from 1to infinity of n^2* e^-2n

Mathematics

1 answer:

Naily [24]2 years ago

3 0

<span>∑ from 1 to infinity of n^2* e^(-2n) = (e^2 + e^4)/(e^2 - 1)^3 = 0.23768</span>

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<h2>Answer = </h2><h2> $= \frac{1}{y {}^{5} x {}^{2} }$ </h2>

Step-by-step explanation:

Greetings !

simplify for variable y first

apply exponent rule

$a {}^{b} .a {}^{c} = a {}^{b +c \\ } \\ y {}^{ - 8}y {}^{5} = y {}^{ - 8 + 3}$

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