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

Answer:

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

a)

$(1+\sqrt{2})^5=\sum_{k=0}^5 \binom{5}{k}1^{5-k}\sqrt{2}^k=\sum_{k=0}^5 \binom{5}{k}2^{\frac{k}{2}}\\=\binom{5}{0}2^{\frac{0}{2}}+\binom{5}{1}2^{\frac{1}{2}}+\binom{5}{2}2^{\frac{2}{2}}+\binom{5}{3}2^{\frac{3}{2}}+\binom{5}{4}2^{\frac{4}{2}}+\binom{5}{5}2^{\frac{5}{2}}\\=1+5\sqrt{2}+10*2+10*2^{\frac{3}{2}}+5*4+1*2^{\frac{5}{2}}\\=41+5\sqrt{2}+10*2^{\frac{3}{2}}+2^{\frac{5}{2}}$

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)

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

e)

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