I love being inspired! 6

The above problem can be written as,

$\displaystyle\sum_{n=1}^{\infty} (n)^{2}.(n+1)^2.x^{n}$

We all know that,

$\displaystyle\sum_{n=1}^{\infty} x^{n}=\frac{1}{1-x}$

Now differentiating with respect to $x$ .

$\displaystyle\sum_{n=1}^{\infty} nx^{n-1}=\frac{1}{(1-x)^{2}}$

Mulipying by $x$ and repeating the same process again and again till we achieve the given expression,

We will end up with,

$\displaystyle\sum_{n=1}^{\infty} (n)^{2}.(n+1)^2.x^{n}=-\frac{4x(x^{2}+4x+1)}{(x-1)^{5}}$

Now we just need to plug-in the value $x=\frac{1}{10}$

And we will get the answer as $\frac{18800}{19683}$

I love being inspired! 5 seems to be in algebra but I'm not sure why this isn't, so I'll just post an algebraic method which is similar to my other solution in part 5.

Firstly, the series converges by the ratio test.

In the following steps, we will subtract $\frac {1}{10}$ of the series we are trying to find the value of, from itself. This reduces the degree of the numerator by 1, and we hope to get an infinite G.P. which we already have the formula for.

Let $W_n=\frac {n^2(n+1)^2}{10^n},W=\displaystyle \sum_{n=1}^\infty W_n$

Note that $W_{n+1}-\frac {1}{10} W_n=4 \frac {(n+1)^3}{10^{n+1}}$ $\Rightarrow \frac {9}{10} W = W_1 + \displaystyle \sum_{n=1}^\infty (W_{n+1}-\frac {1}{10} W_n) =W_1+4 \displaystyle \sum_{n=1}^\infty \frac {(n+1)^3}{10^{n+1}}$

Let $X_n=\frac {(n+1)^3}{10^{n+1}},X=\displaystyle \sum_{n=1}^\infty X_n$

Note that $X_{n+1}-\frac {1}{10} X_n=\frac {3n^2+9n+7}{10^{n+2}}$ $\Rightarrow \frac {9}{10} X = X_1 + \displaystyle \sum_{n=1}^\infty (X_{n+1}-\frac {1}{10} X_n) =X_1+\displaystyle \sum_{n=1}^\infty \frac {3n^2+9n+7}{10^{n+2}}$

Let $Y_n=\frac {3n^2+9n+7}{10^{n+2}},Y=\displaystyle \sum_{n=1}^\infty Y_n$

Note that $Y_{n+1}-\frac {1}{10} Y_n=6 \frac {n+2}{10^{n+3}}$ $\Rightarrow \frac {9}{10} Y = Y_1 + \displaystyle \sum_{n=1}^\infty (Y_{n+1}-\frac {1}{10} Y_n) =Y_1+6 \displaystyle \sum_{n=1}^\infty \frac {n+2}{10^{n+3}}$

Let $Z_n=\frac {n+2}{10^{n+3}},Z=\displaystyle \sum_{n=1}^\infty Z_n$

Note that $Z_{n+1}-\frac {1}{10} Z_n=\frac {1}{10^{n+4}}$ $\Rightarrow \frac {9}{10} Z = Z_1 + \displaystyle \sum_{n=1}^\infty (Z_{n+1}-\frac {1}{10} Z_n) =Z_1+\displaystyle \sum_{n=1}^\infty \frac {1}{10^{n+4}}$

By infinite G.P. formula, $\displaystyle \sum_{n=1}^\infty \frac {1}{10^{n+4}} = \frac {1}{90000}$ .

Working backwards, we get:

$Z=\frac {10}{9}(\frac {3}{10000}+\frac {1}{90000}) = \frac {7}{20250}$

$Y=\frac {10}{9}(\frac {19}{1000}+6*\frac {7}{20250}) = \frac {569}{24300}$

$X=\frac {10}{9}(\frac {8}{100}+\frac {569}{24300}) = \frac {2513}{21870}$

$W=\frac {10}{9}(\frac {4}{10}+4*\frac {2513}{21870}) = \frac {18800}{19683}$

$A+B = \boxed{38483}$

(I agree a calculus solution is more elegant though)

$\frac{18800}{19683}$

Answer: $\boxed{38483}$