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3 years ago

In 0.95 in which place is in the 5

Mathematics

1 answer:

svlad2 [7]3 years ago

5 0

Answer: hundredth

Step-by-step explanation:

You have units tenths then hundredths

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<img src="https://tex.z-dn.net/?f=Find%20%7D%20%5Csum%20r%5E%7B2%7D%20%5Ccdot%7B%20%7D%5E%7B20%7D%20C_%7Br%7D%5Cend%7Bequation%7

Serggg [28]

I bet the sum you're referring to is supposed to be

$\displaystyle \sum_{r=0}^{20} r^2 \times {}^{20}C_r$

or equivalently,

$\displaystyle \sum_{r=0}^{20} r^2 \binom{20}r$

where $\binom nk = \frac{n!}{k!(n-k)!}$ is the binomial coefficient.

Recall the binomial series,

$(1+x)^\alpha = \displaystyle \sum_{r=0}^\infty \binom\alpha r x^r$

which is valid for |x| < 1. (Note that if r > α, the binomial coefficient is defined to be zero, so there really are only α many terms when α is a whole number.)

Differentiating both sides with respect to x gives

$\alpha (1+x)^{\alpha-1} = \displaystyle \sum_{r=0}^\infty r \binom\alpha r x^{r-1}$

Multiply both sides by some arbitrary x in |x| < 1 :

$\alpha x (1+x)^{\alpha-1} = \displaystyle \sum_{r=0}^\infty r \binom\alpha r x^r$

Repeat:

$\alpha (1+x)^{\alpha-1} + \alpha(\alpha-1) x(1+x)^{\alpha-2} = \displaystyle \sum_{r=0}^\infty r^2 \binom\alpha r x^{r-1}$

$\alpha x (1+x)^{\alpha-1} + \alpha(\alpha-1) x^2 (1+x)^{\alpha-2} = \displaystyle \sum_{r=0}^\infty r^2 \binom\alpha r x^r$

Let α = 20, and let x approach 1 from below. The right side converges to the sum we want, while the left side converges to

$20 \times 2^{19} + 20\times19\times 2^{18} = (20 + 10\times19)\times2^{19} = \boxed{210\times2^{19}}$

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$-6x -2x -x = 6$
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$9x=6$
$x= \frac{6}{9}= \frac{2}{3}$

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