sum from n=3 to infinity of 1/(4^n)

Lösung

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

\frac{1}{48}

Dezimale

0.02083 \dots

Schritte zur Lösung

n = 3 \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}} - \frac{1}{4 ^{1}} - \frac{1}{4 ^{2}}

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

= \frac{4}{3} - \frac{1}{4 ^{0}} - \frac{1}{4 ^{1}} - \frac{1}{4 ^{2}}

Vereinfache

\frac{4}{3} - \frac{1}{4 ^{0}} - \frac{1}{4 ^{1}} - \frac{1}{4 ^{2}} : \frac{1}{48}

= \frac{1}{48}

\int \frac{1}{5 x ^{3}} d x

d er i v a t i v e \frac{1}{5} x^{15}

\int \frac{x ^{2}}{( x ^{2} + 4 ) ^{2}} d x

s l o p e (- 3, 4), (2, - 2)

\frac{\partial}{\partial x} (2 y + 2 x^{2})