Question

Use the binomial theorem to expand the following binomial expressions.

a. (1 + √2)^5
b. (1 + i)^9
c. (1 − π)^5 (Hint: 1 − π = 1 + (−π).)
d. (√2 + i)^6
e. (2 − i)^6

Answer 1

Remember, the expansion of
`(x+y)^n` is
`(x+y)^n=\sum_(k=0)^n \binom{n}{k}x^(n-k)y^k`, where
`\binom{n}{k}=(n!)/((n-k)!k!)`.

a)

`(1+√(2))^5=\sum_(k=0)^5 \binom{5}{k}1^(5-k)√(2)^k=\sum_(k=0)^5 \binom{5}{k}2^{(k)/(2)}\\=\binom{5}{0}2^{(0)/(2)}+\binom{5}{1}2^{(1)/(2)}+\binom{5}{2}2^{(2)/(2)}+\binom{5}{3}2^{(3)/(2)}+\binom{5}{4}2^{(4)/(2)}+\binom{5}{5}2^{(5)/(2)}\\=1+5√(2)+10*2+10*2^{(3)/(2)}+5*4+1*2^{(5)/(2)}\\=41+5√(2)+10*2^{(3)/(2)}+2^{(5)/(2)}`

b)

`(1+i)^9=\sum_(k=0)^9 \binom{9}{k}1^(9-k)i^k=\sum_(k=0)^9 \binom{9}{k}i^k\\=\binom{9}{0}i^0+\binom{9}{1}i^1+\binom{9}{2}i^2+\binom{9}{3}i^3+\binom{9}{4}i^4+\binom{9}{5}i^5+\binom{9}{6}i^6+\\+\binom{9}{7}i^7+\binom{9}{8}i^8+\binom{9}{9}i^9\\=1+9i-36-84i+126+126i-+84-36i+9+i\\=16+16i`

c)

`(1-\pi)^5=\sum_(k=0)^5 \binom{5}{k}1^(9-k)(-\pi)^k=\sum_(k=0)^5 \binom{5}{k}(-\pi)^k\\=\binom{5}{0}(-\pi)^0+\binom{5}{1}(-\pi)^1+\binom{5}{2}(-\pi)^2+\binom{5}{3}(-\pi)^3+\binom{5}{4}(-\pi)^4+\binom{5}{5}(-\pi)^5\\=1-5+10\pi^2-10\pi^3+5\pi^4-\pi^5`

d)

`(√(2)+i)^6=\sum_(k=0)^6 \binom{6}{k}√(2)^(6-k)i^k\\=\binom{6}{0}√(2)^(6)i^0+\binom{6}{1}√(2)^(5)i+\binom{6}{2}√(2)^(4)i^2+\binom{6}{3}√(2)^(3)i^3+\binom{6}{4}√(2)^(2)i^4+\binom{6}{5}√(2)i^5+\binom{6}{6}√(2)^(0)i^6`

e)

`(2-i)^6=\sum_(k=0)^6 \binom{6}{k}2^(6-k)(-i)^k\\=\binom{6}{0}2^(6)(-i)^0+\binom{6}{1}2^(5)(-i)^1+\binom{6}{2}2^(4)(-i)^2+\binom{6}{3}2^(3)(-i)^3+\binom{6}{4}2^(2)(-i)^4+\binom{6}{5}2^(1)(-i)^5+\binom{6}{k}2^(0)(-i)^6\\=1-32i+\binom{6}{2}16i^2-\binom{6}{3}8i^3+\binom{6}{4}4i^4-\binom{6}{5}2i^5+\binom{6}{k}i^6`