sum from n=1 to infinity of 1/(4^n)

解答

n = 1 \sum \infty \frac{1}{4 ^{n}}

\frac{1}{3}

十进制

0.33333 \dots

求解步骤

n = 1 \sum \infty \frac{1}{4 ^{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 \frac{1}{4 ^{n}} - \frac{1}{4 ^{0}}

n = 0 \sum \infty \frac{1}{4 ^{n}} = \frac{4}{3}

= \frac{4}{3} - \frac{1}{4 ^{0}}

化简

\frac{4}{3} - \frac{1}{4 ^{0}} : \frac{1}{3}

= \frac{1}{3}

l a pl a ce t r an s f or m (t^{2} - 3 t + 2) sin (3 t)

d er i v a t i v e y = sin (9 x) + cos (2 x)

\frac{d}{d x} (3 arcsin (\frac{x}{2}))

\int 2 sin (3 x) d x

y^{^{'''}} - y^{^{'}} = sin (x)