Question

`\large{\bold \red{ \sum \limits_(8)^(4) {x}^(2) + 9( \frac{ \cos^(2)\infty }{ { \sin}^(2) \infty } )}}`

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Answer 1

No solution

Explanation:

We have

`$\sum_(x=8)^(4)x^2 + 9\left((\cos^2(\infty))/(\sin^2(\infty)) \right)$`

For the sum it is not correct to assume

`$\sum_(x=8)^(4)x^2= 8^2 + 7^2+6^2+5^2+4^2 = 64+49+36+25+16 = 190$`

Note that for

`$\sum_(x=a)^b f(x)$`

it is assumed
`a\leq x \leq b` and in your case
`\\exists x\in\mathbb{Z}: a\leq x\leq b` for
`a>b`

In fact, considering a set
`S` we have

`$\sum_(x=a)^b (S \cup \varnothing) = \sum_(x=a)^b S + \sum_(x=a)^b \varnothing $` that satisfy
`S = S \cup \varnothing`

This means that, by definition
`\sum_(x=a)^b \varnothing = 0`

Therefore,

`$\sum_(x=8)^(4)x^2 = 0$`

because the sum is empty.

For

`9\left((\cos^2(\infty))/(\sin^2(\infty)) \right)`

we have other problems. Actually, this case is really bad.

Note that
`\cos^2(\infty)` has no value. In fact, if we consider for the case

`$\lim_(x \to \infty) \cos^2(x)$`, the cosine function oscillates between
`[-1, 1]` , and therefore it is undefined. Thus, we cannot evaluate

`9\left((\cos^2(\infty))/(\sin^2(\infty)) \right)`

and then

`$\sum_(x=8)^(4)x^2 + 9\left((\cos^2(\infty))/(\sin^2(\infty)) \right)$`

has no solution