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3 years ago

Directions (0.13 to 0.16): Find the odd one out.

Mathematics

1 answer:

Alja [10]3 years ago

4 0

Answer:

$uttar \: pradesh$

Step-by-step explanation:

$because \: all \: except \: uttar \: pradesh \\ \: have \: sea \: coast \\ \: \\need \: brainliest \: and \: thankss...$

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X - 7 < -20 inequalify

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Step-by-step explanation:

x - 7 < -20

x < -20 + 7

x < -13

X = (-12, -11, -10,...)

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2 years ago

Can you tell me which statements go in "agree" and "disagree"?

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Step-by-step explanation:

wefasv

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3 years ago

A group of 4 people are sharing jellybeans. Each person wants 6 jellybeans and each box has 8 jellybeans.

loris [4]

Answer:

3

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2 years ago

Which expression is equivalent to 8/9 divided by 3/4

ipn [44]

Step-by-step explanation:

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3 years ago

Read 2 more answers

Find a compact form for generating functions of the sequence 1, 8,27,... , k^3

pantera1 [17]

This sequence has generating function

$F(x)=\displaystyle\sum_{k\ge0}k^3x^k$

(if we include $k=0$ for a moment)

Recall that for $|x|$ , we have

$\displaystyle\frac1{1-x}=\sum_{k\ge0}x^k$

Take the derivative to get

$\displaystyle\frac1{(1-x)^2}=\sum_{k\ge0}kx^{k-1}=\frac1x\sum_{k\ge0}kx^k$

$\implies\dfrac x{(1-x)^2}=\displaystyle\sum_{k\ge0}kx^k$

Take the derivative again:

$\displaystyle\frac{(1-x)^2+2x(1-x)}{(1-x)^4}=\sum_{k\ge0}k^2x^{k-1}=\frac1x\sum_{k\ge0}k^2x^k$

$\implies\displaystyle\frac{x+x^2}{(1-x)^3}=\sum_{k\ge0}k^2x^k$

Take the derivative one more time:

$\displaystyle\frac{(1+2x)(1-x)^3+3(x+x^2)(1-x)^2}{(1-x)^6}=\sum_{k\ge0}k^3x^{k-1}=\frac1x\sum_{k\ge0}k^3x^k$

$\implies\displaystyle\frac{x+4x^3+x^3}{(1-x)^4}=\sum_{k\ge0}k^3x^k$

so we have

$\boxed{F(x)=\dfrac{x+4x^3+x^3}{(1-x)^4}}$

5 0

3 years ago