espandere (7a+2n)^4

Soluzione

espandere

(7 a + 2 n)^{4}

2401 a^{4} + 2744 a^{3} n + 1176 a^{2} n^{2} + 224 a n^{3} + 16 n^{4}

Fasi della soluzione

(7 a + 2 n)^{4}

Applicare il teorema binomiale:

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

a = 7 a, b = 2 n

= i = 0 \sum 4 (i 4) (7 a)^{(4 - i)} (2 n)^{i}

Espandere la sommatoria

= \frac{4 !}{0 ! ( 4 - 0 ) !} (7 a)^{4} (2 n)^{0} + \frac{4 !}{1 ! ( 4 - 1 ) !} (7 a)^{3} (2 n)^{1} + \frac{4 !}{2 ! ( 4 - 2 ) !} (7 a)^{2} (2 n)^{2} + \frac{4 !}{3 ! ( 4 - 3 ) !} (7 a)^{1} (2 n)^{3} + \frac{4 !}{4 ! ( 4 - 4 ) !} (7 a)^{0} (2 n)^{4}

Semplificare

\frac{4 !}{0 ! ( 4 - 0 ) !} (7 a)^{4} (2 n)^{0} : 2401 a^{4}

Semplificare

\frac{4 !}{1 ! ( 4 - 1 ) !} (7 a)^{3} (2 n)^{1} : 2744 a^{3} n

Semplificare

\frac{4 !}{2 ! ( 4 - 2 ) !} (7 a)^{2} (2 n)^{2} : 1176 a^{2} n^{2}

Semplificare

\frac{4 !}{3 ! ( 4 - 3 ) !} (7 a)^{1} (2 n)^{3} : 224 a n^{3}

Semplificare

\frac{4 !}{4 ! ( 4 - 4 ) !} (7 a)^{0} (2 n)^{4} : 16 n^{4}

= 2401 a^{4} + 2744 a^{3} n + 1176 a^{2} n^{2} + 224 a n^{3} + 16 n^{4}

s im pl i f y 2 \cdot (- 0.5)

s im pl i f y 5 a b^{2} + 2 a b^{2}

5 \frac{1}{6}

s im pl i f y 5.76

s im pl i f y 9 \cdot 1 0^{- 2}