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Natasha2012 [34]
3 years ago
14

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

Mathematics
1 answer:
Mkey [24]3 years ago
4 0

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

2<em>k</em> - 5 = 3   ==>   2<em>k</em> = 8   ==>   <em>k</em> = 4

so that the coefficient would be

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

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Answer:

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Step-by-step explanation:

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The probability mass function of <em>X</em> is:

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