sum from n=1 to infinity of (sin(3))^n

解答

n = 1 \sum \infty (sin (3))^{n}

\frac{1}{1 - sin ( 3 )} - 1

十进制

0.16430 \dots

求解步骤

n = 1 \sum \infty (sin (3))^{n}

n = k \sum \infty (a_{n}) = n = 0 \sum \infty (a_{n}) - a_{0} - a_{1} - \dots a_{k - 1}

= n = 0 \sum \infty (sin (3))^{n} - (sin (3))^{0}

n = 0 \sum \infty (sin (3))^{n} = \frac{1}{1 - sin ( 3 )}

= \frac{1}{1 - sin ( 3 )} - (sin (3))^{0}

化简

= \frac{1}{1 - sin ( 3 )} - 1

\frac{d}{d x} (e^{x + y} - 1)

\int \frac{sin ^{4} ( x )}{cos ^{6} ( x )} d x

t an g e n t f (x) = 7 x, at x = 4

n = 5 \sum \infty \frac{n ^{3}}{2 ^{n}}

f^{^{''}} (x) = 2