desarrollar (3e+4f)^8

Solución

desarrollar

(3 e + 4 f)^{8}

65536 f^{8} + 393216 e f^{7} + 1032192 e^{2} f^{6} + 1548288 e^{3} f^{5} + 1451520 e^{4} f^{4} + 870912 e^{5} f^{3} + 326592 e^{6} f^{2} + 69984 e^{7} f + 6561 e^{8}

Pasos de solución

(3 e + 4 f)^{8}

Aplicar el teorema del binomio:

(a + b)^{n} = i = 0 \sum n (i n) a^{(n - i)} b^{i}

a = 3 e, b = 4 f

= i = 0 \sum 8 (i 8) (3 e)^{(8 - i)} (4 f)^{i}

Expandir sumatorio

= \frac{8 !}{0 ! ( 8 - 0 ) !} (3 e)^{8} (4 f)^{0} + \frac{8 !}{1 ! ( 8 - 1 ) !} (3 e)^{7} (4 f)^{1} + \frac{8 !}{2 ! ( 8 - 2 ) !} (3 e)^{6} (4 f)^{2} + \frac{8 !}{3 ! ( 8 - 3 ) !} (3 e)^{5} (4 f)^{3} + \frac{8 !}{4 ! ( 8 - 4 ) !} (3 e)^{4} (4 f)^{4} + \frac{8 !}{5 ! ( 8 - 5 ) !} (3 e)^{3} (4 f)^{5} + \frac{8 !}{6 ! ( 8 - 6 ) !} (3 e)^{2} (4 f)^{6} + \frac{8 !}{7 ! ( 8 - 7 ) !} (3 e)^{1} (4 f)^{7} + \frac{8 !}{8 ! ( 8 - 8 ) !} (3 e)^{0} (4 f)^{8}

\frac{8 !}{0 ! ( 8 - 0 ) !} (3 e)^{8} (4 f)^{0} = 6561 e^{8}

Simplificar

\frac{8 !}{1 ! ( 8 - 1 ) !} (3 e)^{7} (4 f)^{1} : 69984 e^{7} f

Simplificar

\frac{8 !}{2 ! ( 8 - 2 ) !} (3 e)^{6} (4 f)^{2} : 326592 e^{6} f^{2}

Simplificar

\frac{8 !}{3 ! ( 8 - 3 ) !} (3 e)^{5} (4 f)^{3} : 870912 e^{5} f^{3}

Simplificar

\frac{8 !}{4 ! ( 8 - 4 ) !} (3 e)^{4} (4 f)^{4} : 1451520 e^{4} f^{4}

Simplificar

\frac{8 !}{5 ! ( 8 - 5 ) !} (3 e)^{3} (4 f)^{5} : 1548288 e^{3} f^{5}

Simplificar

\frac{8 !}{6 ! ( 8 - 6 ) !} (3 e)^{2} (4 f)^{6} : 1032192 e^{2} f^{6}

Simplificar

\frac{8 !}{7 ! ( 8 - 7 ) !} (3 e)^{1} (4 f)^{7} : 393216 e f^{7}

Simplificar

\frac{8 !}{8 ! ( 8 - 8 ) !} (3 e)^{0} (4 f)^{8} : 65536 f^{8}

= 6561 e^{8} + 69984 e^{7} f + 326592 e^{6} f^{2} + 870912 e^{5} f^{3} + 1451520 e^{4} f^{4} + 1548288 e^{3} f^{5} + 1032192 e^{2} f^{6} + 393216 e f^{7} + 65536 f^{8}

Reescribir en la forma estándar

= 65536 f^{8} + 393216 e f^{7} + 1032192 e^{2} f^{6} + 1548288 e^{3} f^{5} + 1451520 e^{4} f^{4} + 870912 e^{5} f^{3} + 326592 e^{6} f^{2} + 69984 e^{7} f + 6561 e^{8}

e x p an d P_{1} (6, 1) P_{2} (1, - 4)

e x p an d \frac{1}{s ( s + 1 ) ^{3}}

s im pl i f y (x + 2 + 2 i) \cdot (x + 2 i) \cdot (x - 1 + i)

e x p an d (x - 9) (x - 3)

e x p an d \frac{ln ( x + 1 ) - ln ( x ) ln ( ( x + 1 ) x )}{ln ( ( x + 1 ) x )}