If $$S(x) = (1+x) + 2(1+x)^2 + 3(1+x)^3 + \cdots + 60(1+x)^{60}$$, $$x \neq 0$$, and $$(60)^2 S(60) = a(b)^b + b$$, where $$a, b \in N$$, then $$(a + b)$$ equal to ___________

We have $$S(x) = \sum_{k=1}^{60} k(1+x)^k$$ and need to find $$(a+b)$$ where $$(60)^2 S(60) = a(b)^b + b$$.

The standard result for the sum is $$\sum_{k=1}^{n} k r^k = \frac{r(1 - (n+1)r^n + nr^{n+1})}{(1-r)^2}$$.

In our case $$r = (1+x)$$ and $$n = 60$$, so with $$x = 60$$ we have $$r = 61$$.

Substituting into the formula gives $$S(60) = \frac{61(1 - 61\cdot61^{60} + 60\cdot61^{61})}{(1-61)^2} = \frac{61(1 - 61^{61} + 60\cdot61^{61})}{3600}$$.

This simplifies to $$S(60) = \frac{61(1 + 59\cdot61^{61})}{3600}$$.

Multiplying by $$(60)^2 = 3600$$ yields $$(60)^2 S(60) = 3600\cdot S(60) = 61(1 + 59\cdot61^{61}) = 61 + 59\cdot61^{62}$$.

Comparing with the form $$a(b)^b + b$$ shows that $$59\cdot61^{62} + 61 = a\cdot b^b + b$$.

Choosing $$b = 61$$ gives $$a\cdot61^{61} + 61 = 59\cdot61^{62} + 61$$, so $$a\cdot61^{61} = 59\cdot61^{62}$$ and hence $$a = 59\cdot61 = 3599$$.

It follows that $$a+b = 3599 + 61 = 3660$$.

The answer is 3660.