|
- algebra precalculus - Evaluating $\frac {1} {a^ {2025}}+\frac {1} {b . . .
When I tried to solve this problem, I found a solution (official) video on YouTube That is a = −b, c = 2024 a = b, c = 2024 and the correct answer is 1 20242025 1 2024 2025 Is there an alternative solution but not using (a + b)(a + c)(b + c) + abc = (a + b + c)(ab + ac + bc) (a + b) (a + c) (b + c) + a b c = (a + b + c) (a b + a c + b c) ?
- Evaluating $\\sqrt{119^2+120^2}$ with clever algebra
Numbers $(119,120,169)$ are Pythagorean triples, i e $119^2+120^2=169^2$ I'm wondering is it possible to start from $119^2+120^2$ and get $169^2$ algebraically without evaluating $119^2$ and $120
- calculus - Evaluating $\int_0^ {\pi 2} \frac {\sqrt {\tan x}} {\sin x . . .
Evaluating ∫π 2 0 tan x√ sin x(cos x+sin x) dx ∫ 0 π 2 tan x sin x (cos x + sin x) d x Ask Question Asked 1 year, 11 months ago Modified 7 months ago
- limits - Evaluating $\lim\limits_ {n \to \infty} ( (n^3 + n^2 + n + 1 . . .
Are you familiar with evaluating limits of the form x for ? If so, can you transform the given limit into the form above? – sudeep5221 Oct 20, 2023 at 18:33 @sudeep5221 I'm afraid I don't If were an integer, I would try to use the binomial theorem, but it seems that it can be an arbitrary positive real number
- Evaluating $ \\lim_{x \\to 0} \\frac{e - (1 + 2x)^{1 2x}}{x} $ without . . .
Evaluating limx→0 e−(1+2x)1 2x x lim x → 0 e (1 + 2 x) 1 2 x x without using any expansion series [closed] Ask Question Asked 10 months ago Modified 10 months ago
- Evaluating $\\lim_{n\\to\\infty}\\left( \\frac{\\cos\\frac{\\pi}{2n . . .
The problem is to solve: $$\lim_ {n\to\infty}\left ( \frac {\cos\frac {\pi} {2n}} {n+1}+\frac {\cos\frac {2\pi} {2n}} {n+1 2}+\dots+\frac {\cos\frac {n\pi} {2n}} {n+1
- Evaluating $\\lim\\limits_{n\\to\\infty} e^{-n} \\sum\\limits_{k=0}^{n . . .
I'm supposed to calculate: $$\\lim_{n\\to\\infty} e^{-n} \\sum_{k=0}^{n} \\frac{n^k}{k!}$$ By using WolframAlpha, I might guess that the limit is $\\frac{1}{2
- Evaluating $\\sum_{n=1}^{\\infty} \\frac{1}{n^2+1}$
I evaluated it through wolfram alpha, it gave me 1 2(π coth(π) − 1) 1 2 (π coth (π) 1) What would be a good way to start evaluating this series?
|
|
|