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Lerok [7]
3 years ago
14

Use the binomial theorem to expand the following binomial expressions.

Mathematics
1 answer:
andre [41]3 years ago
5 0

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

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Question:

A question related to this found at brainly (ID:4482275) is stated below.

The area of the right triangle shown is 24 square feet. Which equations can be used to find the lengths of the legs of the triangle? Check all that apply.

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Step-by-step explanation:

Find attached the diagram.

Area of triangle = ½ × base × height

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24 = 0.5(x)(x+2)

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The equations that can be used to find the lengths of the legs of the triangle must be equivalent to 0.5(x)(x + 2) = 24

On expanding this: 0.5(x)(x + 2) = 24

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b) x(x + 2) = 24

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c) x² + 2x – 24 = 0

0.5(x²+2x) = 24

0.5x²+x - 24 = 0 is not equal to x²+2x- 24 = 0

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x² + 2x = 2(24)

x² + 2x – 48 = 0

Correct option (D)

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2(x² + 2x +48)= 0

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Correct (E)

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