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3 years ago

9

X y

1 answer:

mario62 [17]3 years ago

8 0

Let's prove each one:
A) y=x-3
x=0→y=0-3→y=-3=-3 ok
x=2→y=2-3→y=-1 different to 5 No

B) y=x+3
x=0→y=0+3→y=3 different to -3 No

C) y=4x-3
x=0→y=4(0)-3=0-3→y=-3=-3 ok
x=2→y=4(2)-3=8-3→y=5=5 ok
x=3→y=4(3)-3=12-3→y=9=9 ok
x=4→y=4(4)-3=16-3→y=13=13 ok

D) y=4x+3
x=0→y=4(0)+3=0+3→y=3 different to -3 No

Answer: Option C) y=4x-3

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

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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:

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

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

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