Power 5

For $-1< x < 1$ , we have:

$\begin{aligned} \sum_{n=0}^\infty x^n & = \frac 1{1-x} & \small {\color{#3D99F6}\text{Differentiate both sides}} \\ \sum_{n=1}^\infty nx^{n-1} & = \frac 1{(1-x)^2} & \small {\color{#3D99F6}\text{Multiply both sides by }x} \\ \sum_{n=1}^\infty nx^n & = \frac x{(1-x)^2} & \small {\color{#3D99F6}\text{Differentiate again}} \\ \sum_{n=1}^\infty n^2x^{n-1} & = \frac {1+x}{(1-x)^3} & \small {\color{#3D99F6} \times x \text{ again}} \\ \sum_{n=1}^\infty n^2x^n & = \frac {x(1+x)}{(1-x)^3} & \small {\color{#3D99F6}\text{Differentiate again}} \\ \sum_{n=1}^\infty n^3x^{n-1} & = \frac {1+4x+x^2}{(1-x)^4} & \small {\color{#3D99F6} \times x \text{ again}} \\ \sum_{n=1}^\infty n^3x^n & = \frac {x+4x^2+x^3}{(1-x)^4} & \small {\color{#3D99F6}\text{Differentiate again}} \\ \sum_{n=1}^\infty n^4x^{n-1} & = \frac {1+11x+11x^2+x^3}{(1-x)^5} & \small {\color{#3D99F6} \times x \text{ again}} \\ \sum_{n=1}^\infty n^4x^n & = \frac {x+11x^2+11x^3+x^4}{(1-x)^5} & \small {\color{#3D99F6}\text{Differentiate again}} \\ \sum_{n=1}^\infty n^5x^{n-1} & = \frac {1+26x+66x^2+26x^3+x^4}{(1-x)^6} & \small {\color{#3D99F6} \times x \text{ again}} \\ \sum_{n=1}^\infty n^5x^n & = \frac {x+26x^2+66x^3+26x^4+x^5}{(1-x)^6} & \small {\color{#3D99F6} \text{Put }x = \frac 13} \\ \sum_{n=1}^\infty \frac {n^5}{3^n} & = \frac {\frac 13 +\frac {26}{3^2}+\frac {66}{3^3}+\frac {26}{3^4}+\frac 1{3^5}}{\left(1-\frac 13\right)^6} \\ & = \frac {273}4 \end{aligned}$

$\implies a+b = 273+4 = \boxed{277}$ .

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$\text{Infinite sum of GP} \\ \sum x^n = \dfrac{1}{1-x}\\ \dfrac{d}{dx} \sum x^n = \dfrac{d}{dx} \dfrac{1}{1-x} \\ \sum nx^{n-1} = (1-x)^{-2} \\ x\sum nx^{n-1} = x(1-x)^{-2} \\ \sum nx^n = x(1-x)^{-2 } \\ \dfrac{d}{dx} \sum nx^n = \dfrac{d}{dx} x(1-x)^{-2} \\ \sum n^2 x^n = x (1+x)(1-x)^{-3} \\ \text{Similarly, } \\ \sum n^5 x^n = \dfrac{x(x^4+26 x^3+66 x^2+26 x+1))}{(x-1)^6}\\ \boxed{\text{for }x = \dfrac{1}{3}} \\ \sum \dfrac{n^5}{3^n} = \dfrac{\dfrac{1}{3}((\dfrac{1}{3})^4+26 (\dfrac{1}{3})^3+66 (\dfrac{1}{3})^2+26 (\dfrac{1}{3})+1))}{((\dfrac{1}{3})-1)^6}\\ = \dfrac{273}{4} \\ \color{#3D99F6}{\boxed{\therefore a + b = 273 + 4 = 277}}$