Question

Find the scale of relation of the recurring series $2+5x+13x^2+35x^3+.....$

Answer 1

Let $S=2+5x+13x^2+35x^3+...$

Multiplying $S$ with $x$ we get

$xS=2x+5x^2+13x^3+35x^4+...$

$S=2+5x+13x^2+35x^3+...\underset{–}{-xS=2x+5x^2+13x^3+..}(1-x)S=2+3x+8x^2+22x^3+..$

Multiplying $(1-x)S$ with $x$ we get

$x(1-x)S=2x+3x^2+8x^3+22x^4+....$

$(1-x)S=2+3x+8x^2+22x^3+..\underset{–}{-x(1-x)S=2x+3x^2+8x^3+....}{(1-x)}^{2}S=2+x+5x^2+14x^3+...$

Multiplying $(1-x{)}^{2}S$ with $x$ we get

$x(1-x{)}^{2}S=2x+x^2+5x^3+14x^4+...$

${(1-x)}^{2}S=2+x+5x^2+14x^3+...\underset{–}{-x(1-x{)}^{2}S=2x+x^2+5x^3+....}{(1-x)}^{3}S=2-x+4x^2+9x^3+...$

In this series $-1,4,9,...$ forms an AP whose $a_1=-1$ and $d=5$

General term for AP $a_n$$=a_1+(n-1)d=-1+(n-1)5=(5n-6)$

${(1-x)}^{3}S=2-x+4x^2+9x^3+14x^4+.....$

$S^{\prime }=-1+4x+9x^2+14x^3+...$

and $S^{\prime }=-1+4x+9x^2+14x^3+....\infty$ forms an AGP

this tends to infinity

$S=\frac{a}{1-r}+\frac{dr(1-r^{n-1})}{{(1-r)}^2}-\frac{(a+(n-1\left)d\right)r^n}{(1-r)}{S}^{\prime }=\frac{-x}{1-x}+\frac{5x^2(1-x^{n-1})}{{(1-x)}^2}-\frac{[-1+(n-1\left)5\right]x^n}{(1-x)}$

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