From sum to sum

Let the first infinite sum be $S_1$ , therefore,

$\begin{aligned} S_1 & = 1 + \frac{3}{x} + \frac{5}{x^2} + \frac{7}{x^3} + \frac{9}{x^4} + ... \\ & = \sum_{n=0}^\infty (1+2n)\left(\frac{1}{x}\right)^n \end{aligned}$

We note that $S_1$ is an AGP (arithmetic-geometric progression) with first term $a=1$ , common difference $d=2$ and common ratio $r= \dfrac{1}{x}$ and its sum is $S_\infty = \dfrac{a}{1-r} + \dfrac{dr}{(1-r)^2}$ (see wiki ). Therefore, we have:

$\begin{aligned} \frac{1}{1-\frac{1}{x}} +\frac{2\left(\frac{1}{x}\right)}{\left(1-\frac{1}{x}\right)} & = 91 \\ \frac{x}{x-1} + \frac{2x}{(x-1)^2} & = 91 \\ \frac{x(x-1)+2x}{x^2-2x+1} & = 91 \\ 90x^2 - 183x + 91 & = 0 \\ (15x-13)(6x-7) & = 0 \\ \Rightarrow x & = \begin{cases} \frac{13}{15} & \color{#D61F06}{< 1 \quad \text{rejected; for }S_1 \text{ to converge } x>1.} \\ \frac{7}{6} & \color{#3D99F6}{> 1 \quad \text{accepted.}} \end{cases} \end{aligned}$

Let the second infinite sum be $S_2$ , then we have:

$\begin{aligned} S_2 & = 1 + \color{#3D99F6}{\frac{4}{x}} + \color{#D61F06}{\frac{9}{x^2}} + \color{#20A900}{\frac{16}{x^3}} + \color{magenta}{\frac{25}{x^4}} + ... \\ & =1 + \color{#3D99F6}{\frac{3}{x}+\frac{1}{x}} + \color{#D61F06}{\frac{5}{x^2}+\frac{4}{x^2}} + \color{#20A900}{\frac{7}{x^3}+\frac{9}{x^3}} + \color{magenta}{\frac{9}{x^4}+\frac{16}{x^4}} + ... \\ & = 1 + \color{#3D99F6}{\frac{3}{x}} + \color{#D61F06}{\frac{5}{x^2}} + \color{#20A900}{\frac{7}{x^3}} + \color{magenta}{\frac{9}{x^4}} + ...\color{#3D99F6}{\frac{1}{x}} + \color{#D61F06}{\frac{4}{x^2}} + \color{#20A900}{\frac{9}{x^3}} + \color{magenta}{\frac{16}{x^4}} + ... \\ & = 1 + \color{#3D99F6}{\frac{3}{x}} + \color{#D61F06}{\frac{5}{x^2}} + \color{#20A900}{\frac{7}{x^3}} + \color{magenta}{\frac{9}{x^4}} + ... + \frac{1}{x} \left(\color{#3D99F6}{1} + \color{#D61F06}{\frac{4}{x}} + \color{#20A900}{\frac{9}{x^2}} + \color{magenta}{\frac{16}{x^3}} + ... \right) \\ & = S_1 + \frac{1}{x}S_2 \\ \Rightarrow S_2 & = S_1 + \frac{6}{7}S_2 \\ S_2 & = 7S_1 = \boxed{637} \end{aligned}$

Let

$\begin{aligned}S=&\ 1+\dfrac4x+\dfrac9{x^2}+\dfrac{16}{x^3}+\dfrac{25}{x^4}+\cdots\Rightarrow(1)\end{aligned}$

Multiplying with $\dfrac1x$ in equation $(1)$

$\begin{aligned}\dfrac1xS=&\ \dfrac1x+\dfrac4{x^2}+\dfrac9{x^3}+\dfrac{16}{x^4}+\dfrac{25}{x^5}+\cdots\Rightarrow(2)\end{aligned}$

Subtracting equation $(1)$ with equation $(2)$

$\begin{aligned}\left(1-\dfrac1x\right)S=&\ 1+\dfrac{4-1}x+\dfrac{9-4}{x^2}+\dfrac{16-9}{x^3}+\dfrac{25-16}{x^4}+\cdots\\\left(\dfrac{x-1}x\right)S=&\ 1+\dfrac3x+\dfrac5{x^2}+\dfrac7{x^3}+\dfrac9{x^4}+\cdots\\\left(\dfrac{x-1}x\right)S=&\ 91\\S=&\ \dfrac{91x}{x-1}\Rightarrow(3)\end{aligned}$

From Equation,

$\begin{aligned}1+\dfrac3x+\dfrac5{x^2}+\dfrac7{x^3}+\dfrac9{x^4}+\cdots=&\ 91\Rightarrow(4)\end{aligned}$

Multiplying with $\dfrac1x$ in equation $(4)$

$\begin{aligned}\dfrac1x+\dfrac3{x^2}+\dfrac5{x^3}+\dfrac7{x^4}+\dfrac9{x^5}+\cdots=&\ \dfrac{91}x\Rightarrow(5)\end{aligned}$

Subtracting equation $(4)$ with equation $(5)$

$\begin{aligned}1+\dfrac{3-1}x+\dfrac{5-3}{x^2}+\dfrac{7-5}{x^3}+\dfrac{9-7}{x^4}+\cdots=&\ 91-\dfrac{91}x\\\dfrac2x+\dfrac2{x^2}+\dfrac2{x^3}+\dfrac2{x^4}+\cdots=&\ 90-\dfrac{91}x\\\dfrac2{x^2}+\dfrac2{x^3}+\dfrac2{x^4}+\cdots=&\ 90-\dfrac{93}x\\\dfrac{\dfrac2{x^2}}{1-\dfrac1x}=&\ \dfrac{90x-93}x,\ \left|\dfrac1x\right|<1\\\dfrac2{x^2}=&\ \left(\dfrac{90x-93}x\right)\left(\dfrac{x-1}x\right)\\\dfrac2{x^2}=&\ \dfrac{90x^2-183x+93}{x^2}\\2=&\ 90x^2-183x+93,\ x\neq0\\90x^2-183x+91=&\ 0\\(15x-13)(6x-7)=&\ 0\\x=&\ \dfrac{13}{15},\ \dfrac76\end{aligned}$

But $\left|\dfrac1x\right|<1$ Thus, $x=\dfrac76$

Instead $x$ with $\dfrac76$ in equation $(3)$ that is,

$\begin{aligned}S=\dfrac{91\left(\dfrac76\right)}{\dfrac76-1}=\dfrac{\dfrac{637}6}{\dfrac16}=\boxed{637}\end{aligned}$

I used calculus!

First, let $z = \dfrac{1}{\sqrt{x}}$ . Then $1 + \frac3 x + \frac 5{x^2} + \cdots = 1 + 3z^2 + 5z^4 + \cdots \\ = \frac{d}{dz}(z+z^3+z^5+\cdots) = \frac{d}{dz}(z\cdot(1+z^2+z^4+\cdots)) \\ = \frac{d}{dz}\frac{z}{1-z^2} = \frac{1+z^2}{(1-z^2)^2} \\ = \frac{x^2+x}{(x-1)^2}.$ This is equal to 91. Solving for $x$ , $x^2 + x = 91(x-1)^2\ \ \ \therefore\ \ \ x = \tfrac76\ \vee\ x = \tfrac{13}{15},$ but the latter solution we reject because that would cause a divergent series.

Now let $u = 1/x$ . Then $1 + \tfrac 4 x + \frac 9{x^2} + \cdots = 1 + 4u + 9u^2 + \cdots \\ = \frac{d}{du}(u + 2u^2 + 3u^3 + \cdots) = \frac{d}{du}(u\cdot(1 + 2u + 3u^2 + \cdots)) \\ = \frac{d}{du}\left(u\frac{d}{du}(u + u^2 + u^3 + \cdots)\right) = \frac{d}{du}\left(u\frac{d}{du}\frac{u}{1-u}\right) \\ = \frac{d}{du}\left(u\frac{1}{(1-u)^2}\right) = \frac{d}{du}\frac{u}{(1-u)^2} \\ = \frac{1+u}{(1-u)^3} = \frac{x^3+x^2}{(x-1)^3} \\ = \frac{x}{x-1}\cdot \frac{x^2+x}{(x-1)^2} = \frac{7/6}{7/6-1}\cdot 91 \\ = \frac{7}{7-6}\cdot 91 = \boxed{637}.$

Observing that
$1=1$
$4=1+3$
$9=1+3+5$
...
I expressed $\displaystyle1+\frac 4x+\frac 9{x^2}+...$
$\displaystyle=(1+\frac 3x+\frac 5{x^2}+...)(1+\frac 1x+\frac 1{x^2}+...)$
$\displaystyle=(91)(\frac 1{1-\frac 1x})$
$\displaystyle=\frac {91x}{x-1}$
My approach to find $x$ is the same as Ikkyu San's one.

Define $f(y)=\sum_{i=0}^{\infty} y^i = 1/(1-x)$ Then, by differentiating the two equivalent expressions, we get $f'(y) = \sum_{i=0}^\infty (i+1)y^i = f(y)^2$ $f''(y) = \sum_{i=0}^\infty (i+1)(i+2)y^i = 2 f(y)^3$

Let $z=f(1/x)$ , then we know that $2z^2-z-91=0$ $z=7 or -6.5$ $z$ is clearly positive, so we accept the value $z=7$ . We want to calculate

$\sum _{ i=0 }^{ \infty }{ (i+1)^{ 2 } } { x }^{ -i }\quad =\quad 2{ z }^{ 3 }-{ z }^{ 2 }\\ \qquad \qquad \qquad =\quad z(2{ z }^{ 2 }-z)\\ \qquad \qquad \qquad =\quad 91z\\ \qquad \qquad \qquad =\quad 637\\ \\$

Sorry for the formatting, I can't figure out how to get a LaTeX \align to work.

x = 7/ 6

Sum wanted = 637

$1+\frac{3}{x}+ \frac{5}{x^2}+...=91 \quad$ Subtract 1 and multiply by $x$ on both sides

$3+ \frac{5}{x}+ \frac{7}{x^2}+...=90x \quad$ Subtract the first line from the second

$2(1+ \frac{1}{x}+ \frac{1}{x^2}+...)=90x-91 \quad$ This converges as $|x|>1$ . Assume that is the case and multiply by $x-1$

$2x =(90x-91)(x-1)$

$0=90x^2-183x+91$

$x=(183\pm27)/180$

$x=\frac{7}{6} \vee x=\frac{13}{15} \quad$ The latter value for x is not a valid solution, because we assumed $|x|>1$ .

Let's dub the expression to evaluate $t$ :

$t=1+ \frac{4}{x}+ \frac{16}{x^2}+...$

$\frac{t}{x}=\frac{1}{x}+ \frac{4}{x^2}+ \frac{16}{x^3}+...$ Subtract from the line above

$t(1- \frac{1}{x})=1+ \frac{3}{x}+ \frac{5}{x^2}+...=91$

$t=\frac{91}{1- \frac{6}{7}}=91×7=\boxed{637}$

We set $y=\dfrac1x$ . So $1+3y+5y^2+7y^3+9y^4+\cdots=91\quad\Leftrightarrow\quad\sum_{n=1}^\infty(2n-1)y^{n-1} =91.$ Using derivative, we obtain $\sum_{n=1}^\infty(2n-1)y^{n-1} =2\sum_{n=1}^\infty n y^{n-1}-\sum_{n=1}^\infty y^{n-1}=2\sum_{n=1}^\infty(y^n)' -\sum_{n=1}^\infty y^{n-1}=2\Big(\sum_{n=1}^\infty y^n\Big)'-\frac1{1-y}=\frac2{(1-y)^2}-\frac1{1-y},$ i.e. $\frac{1+y}{(1-y)^2}=91\quad\Rightarrow\quad y=\frac67\,\vee\,\, y=\frac{15}{13}.$ However, we reject $\dfrac{15}{13}$ , because there must be $|y|<1$ for the series to converge. Further we have $1+4y+9y^2+16y^3+25y^4+\cdots=1+(1+3)y+(4+5)y^2+(7+9)y^3+(9+16)y^4+(11+25)y^5+\cdots.$ There follows $a=1+4y+9y^2+16y^3+25y^4+\cdots=(1+3y+5y^2+7y^3+9y^4+11y^5\cdots)+(y+4y^2+9y^3+16y^4+25y^5+\cdots)$ In view of the previous result, and replacing $y=\dfrac67$ , we find $a=91+y\cdot a\,\,\,\Leftrightarrow\,\,\,a=\frac{91}{1-y}\,\,\,\Rightarrow\,\,\,a=637.$