espandere (a+2b)^5

Soluzione

espandere

(a + 2 b)^{5}

a^{5} + 10 a^{4} b + 40 a^{3} b^{2} + 80 a^{2} b^{3} + 80 a b^{4} + 32 b^{5}

Fasi della soluzione

(a + 2 b)^{5}

Applicare il teorema binomiale:

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

a = a, b = 2 b

= i = 0 \sum 5 (i 5) a^{(5 - i)} (2 b)^{i}

Espandere la sommatoria

= \frac{5 !}{0 ! ( 5 - 0 ) !} a^{5} (2 b)^{0} + \frac{5 !}{1 ! ( 5 - 1 ) !} a^{4} (2 b)^{1} + \frac{5 !}{2 ! ( 5 - 2 ) !} a^{3} (2 b)^{2} + \frac{5 !}{3 ! ( 5 - 3 ) !} a^{2} (2 b)^{3} + \frac{5 !}{4 ! ( 5 - 4 ) !} a^{1} (2 b)^{4} + \frac{5 !}{5 ! ( 5 - 5 ) !} a^{0} (2 b)^{5}

Semplificare

\frac{5 !}{0 ! ( 5 - 0 ) !} a^{5} (2 b)^{0} : a^{5}

Semplificare

\frac{5 !}{1 ! ( 5 - 1 ) !} a^{4} (2 b)^{1} : 10 a^{4} b

Semplificare

\frac{5 !}{2 ! ( 5 - 2 ) !} a^{3} (2 b)^{2} : 40 a^{3} b^{2}

Semplificare

\frac{5 !}{3 ! ( 5 - 3 ) !} a^{2} (2 b)^{3} : 80 a^{2} b^{3}

Semplificare

\frac{5 !}{4 ! ( 5 - 4 ) !} a^{1} (2 b)^{4} : 80 a b^{4}

Semplificare

\frac{5 !}{5 ! ( 5 - 5 ) !} a^{0} (2 b)^{5} : 32 b^{5}

= a^{5} + 10 a^{4} b + 40 a^{3} b^{2} + 80 a^{2} b^{3} + 80 a b^{4} + 32 b^{5}

s im pl i f y \frac{( x ^{n} - 1 )}{( x - 1 )}

f (x) = arcsin (\frac{( 1 + x ^{2} )}{2})

s im pl i f y \frac{2 ^{- 3} x ^{5} y ^{6} m ^{13}}{2 x ^{5} y ^{- 6} d ^{- 1}}

f a c t or 10 x^{2} \cdot y^{3} - 5 x^{2} \cdot y^{2} + 15 x^{3} \cdot y^{3}

s im pl i f y (\frac{x}{2 - 2 y})^{2}