erweitern (((n+1)(n+2))/2)^2

Lösung

erweitern

(\frac{( n + 1 ) ( n + 2 )}{2})^{2}

\frac{n ^{4}}{4} + \frac{3 n ^{3}}{2} + \frac{13 n ^{2}}{4} + 3 n + 1

Schritte zur Lösung

(\frac{( n + 1 ) ( n + 2 )}{2})^{2}

\frac{( n + 1 ) ( n + 2 )}{2} = \frac{n ^{2} + 3 n + 2}{2}

= (\frac{n ^{2} + 3 n + 2}{2})^{2}

Wende Exponentenregel an:

(\frac{a}{b})^{c} = \frac{a ^{c}}{b ^{c}}

= \frac{( n ^{2} + 3 n + 2 ) ^{2}}{2 ^{2}}

(n^{2} + 3 n + 2)^{2} = n^{4} + 6 n^{3} + 13 n^{2} + 12 n + 4

= \frac{n ^{4} + 6 n ^{3} + 13 n ^{2} + 12 n + 4}{2 ^{2}}

2^{2} = 4

= \frac{n ^{4} + 6 n ^{3} + 13 n ^{2} + 12 n + 4}{4}

Wende Bruchregel an:

\frac{a \pm b}{c} = \frac{a}{c} \pm \frac{b}{c}

\frac{n ^{4} + 6 n ^{3} + 13 n ^{2} + 12 n + 4}{4} = \frac{n ^{4}}{4} + \frac{6 n ^{3}}{4} + \frac{13 n ^{2}}{4} + \frac{12 n}{4} + \frac{4}{4}

= \frac{n ^{4}}{4} + \frac{6 n ^{3}}{4} + \frac{13 n ^{2}}{4} + \frac{12 n}{4} + \frac{4}{4}

Wende Regel an

\frac{a}{a} = 1

\frac{4}{4} = 1

= \frac{n ^{4}}{4} + \frac{6 n ^{3}}{4} + \frac{13 n ^{2}}{4} + \frac{12 n}{4} + 1

Streiche

\frac{6 n ^{3}}{4} : \frac{3 n ^{3}}{2}

= \frac{n ^{4}}{4} + \frac{3 n ^{3}}{2} + \frac{13 n ^{2}}{4} + \frac{12 n}{4} + 1

Teile die Zahlen:

\frac{12}{4} = 3

= \frac{n ^{4}}{4} + \frac{3 n ^{3}}{2} + \frac{13 n ^{2}}{4} + 3 n + 1

e x p an d x^{\frac{1}{2}} (x^{\frac{3}{2}} - 4) (x^{\frac{1}{2}} + 4)

e x p an d (1 - ln (x)) d y

e x p an d (- \frac{1}{23 , 2 + 0 , 9 K}) \cdot (- 0.9 K^{2} + 53.6 K - 793.6)

e x p an d \frac{1}{( z + 1 ) ( z + 2 )}

s im pl i f y 4 (4 cos (4 x) - sin (x) + (- 4) sin (4 x) cos (x))