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Keith_Richards [23]
3 years ago
10

Pls help will mark brainliest​

Mathematics
1 answer:
Alenkinab [10]3 years ago
5 0

Answer:

jajaja y la novia

Step-by-step explanation:

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Step-by-step explanation:

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Determine the coefficient of x^3 in the expansion of (1 – x)^5(1 + 1/x)^5
Mkey [24]

Notice that

(1 - <em>x</em>)⁵ (1 + 1/<em>x</em>)⁵ = ((1 - <em>x</em>) (1 + 1/<em>x</em>))⁵ = (1 - <em>x</em> + 1/<em>x</em> - 1)⁵ = (1/<em>x</em> - <em>x</em>)⁵

Recall the binomial theorem:

\displaystyle(a+b)^n = \sum_{k=0}^n\binom nk a^{n-k}b^k

Let <em>a</em> = 1/<em>x</em>, <em>b</em> = -<em>x</em>, and <em>n</em> = 5. Then

\displaystyle\left(\frac1x-x\right)^5 = \sum_{k=0}^5\binom5k\left(\frac1x\right)^{5-k}(-x)^k = \sum_{k=0}^5 (-1)^k\binom5k x^{2k-5}

We get an <em>x</em> ³ term for

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so that the coefficient would be

(-1)^4\dbinom54 = \boxed{5}

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Shtirlitz [24]

Answer:

.

Step-by-step explanation:

.

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3 years ago
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