Calculus 2: Sum from n=1 to infinity of (n^4 + ((-1)^n)*n^2) / ((-n)^5 + 3(n^3))

Mathematics

1 answer:

WITCHER [35]3 years ago

5 0

Answer:

(n^4 + ((-1)^n)*n^2) / ((-n)^5 + 3(n^3))

n^4 + -1/ -1^5 + 3n^3

n^4 + -1^n / -1^5 +3^3

n^4/ -1 + 3n^3

n^4/ -1 + 3x1^3

n/-1 + 3n

1/-1+3

n = 2

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