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Express the series 2 + 5x + 9x^2 + 14x^3 + 20x^4 + ... as a function of x.

Hint: The coefficiens 2, 5, 9, 14, 20 ... follow a quadratic formula.

Dec 17, 2019

#1
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I will show that

$$2+5x+9x^2+14x^3+20x^4+...= \frac{1}{(1-x)^3} + \frac{1}{(1-x)^2} = \frac{2-x}{(1-x)^3}$$.

Using the binomial series expansion

$$(1+x)^n=1+nx+\frac{n(n-1)}{2!}x^2+\frac{n(n-1)(n-2)}{3!}x^3+...$$

with $$n=-3\ and\ n=-2$$ we get

$$(1+x)^{-3}=1+(-3)x+\frac{-3(-4)}{2!}x^2+\frac{-3(-4)(-5)}{3!}x^3+\frac{-3(-4)(-5)(-6)}{4!}x^4+...$$

$$=1-3x+6x^2-10x^3+15x^4-...$$

and replacing $$x\ with\ -x$$ in the above we will have

$$\frac{1}{(1-x)^3}=(1-x)^{-3}=1+3x+6x^2+10x+15x^4+...$$

And with $$n=-2,$$

$$(1+x)^{-2}=1+(-2)x+\frac{-2(-3)}{2!}x^2+\frac{-2(-3)(-4)}{3!}x^3+\frac{-2(-3)(-4)(-5)}{4!}x^4+...$$

$$=1-2x+3x^2-4x^3+5x^4-...$$,

and replacing again x with - x, we get

$$\frac{1}{(1-x)^2}=(1-x)^{-2}=1+2x+3x^2+4x^3+5x^4+...$$;

and finally adding the two series for n = -3 and n = -2, we obtain the function I gave above.

Dec 17, 2019