Key Concepts and Formulas

• Arithmetico-Geometric Series: A series where each term is the product of a term from an arithmetic progression and a term from a geometric progression.
• Sum of a Finite Geometric Series: $S_n = \frac{a(1-r^n)}{1-r}$, where $a$ is the first term, $r$ is the common ratio, and $n$ is the number of terms.
• Method for Summing Arithmetico-Geometric Series: Multiply the series by the common ratio of the geometric part and subtract the original series from it to obtain a geometric series.

Step-by-Step Solution

Step 1: Define the series and identify its type. We are given the series $\mathrm{S}(x) = (1+x) + 2(1+x)^2 + 3(1+x)^3 + \cdots + 60(1+x)^{60}$. This is an arithmetico-geometric series where the arithmetic part is $1, 2, 3, \ldots, 60$ and the geometric part has a common ratio of $(1+x)$.

Step 2: Manipulate the series to find a closed-form expression. Let $r = (1+x)$. Then the series is $S(x) = 1 \cdot r + 2 \cdot r^2 + 3 \cdot r^3 + \cdots + 60 \cdot r^{60}$. Multiply $S(x)$ by $r$: $rS(x) = 1 \cdot r^2 + 2 \cdot r^3 + 3 \cdot r^4 + \cdots + 59 \cdot r^{60} + 60 \cdot r^{61}$.

Subtract $rS(x)$ from $S(x)$: $S(x) - rS(x) = (r + 2r^2 + 3r^3 + \cdots + 60r^{60}) - (r^2 + 2r^3 + 3r^4 + \cdots + 59r^{60} + 60r^{61})$ $(1-r)S(x) = r + (2r^2 - r^2) + (3r^3 - 2r^3) + \cdots + (60r^{60} - 59r^{60}) - 60r^{61}$ $(1-r)S(x) = r + r^2 + r^3 + \cdots + r^{60} - 60r^{61}$.

The terms $r + r^2 + r^3 + \cdots + r^{60}$ form a geometric series with first term $a=r$, common ratio $r$, and $n=60$ terms. The sum of this geometric series is $\frac{r(1-r^{60})}{1-r}$.

So, $(1-r)S(x) = \frac{r(1-r^{60})}{1-r} - 60r^{61}$.

Substitute back $r = (1+x)$: $(1 - (1+x))S(x) = \frac{(1+x)(1-(1+x)^{60})}{1-(1+x)} - 60(1+x)^{61}$ $-xS(x) = \frac{(1+x)(1-(1+x)^{60})}{-x} - 60(1+x)^{61}$ $-xS(x) = -\frac{(1+x)(1-(1+x)^{60})}{x} - 60(1+x)^{61}$.

Multiply by $-1$: $xS(x) = \frac{(1+x)(1-(1+x)^{60})}{x} + 60(1+x)^{61}$.

Multiply by $x$: $x^2S(x) = (1+x)(1-(1+x)^{60}) + 60x(1+x)^{61}$ $x^2S(x) = (1+x) - (1+x)^{61} + 60x(1+x)^{61}$.

Step 3: Evaluate the expression at x = 60. We are given $(60)^2 \mathrm{S}(60) = \mathrm{a}(\mathrm{b})^{\mathrm{b}}+\mathrm{b}$. Substitute $x=60$ into the derived expression for $x^2S(x)$: $(60)^2 S(60) = (1+60) - (1+60)^{61} + 60(60)(1+60)^{61}$ $(60)^2 S(60) = 61 - (61)^{61} + 3600(61)^{61}$ $(60)^2 S(60) = 61 + (3600 - 1)(61)^{61}$ $(60)^2 S(60) = 61 + 3599(61)^{61}$.

Rearranging to match the form $a(b)^b + b$: $(60)^2 S(60) = 3599(61)^{61} + 61$.

Step 4: Identify the values of a and b and calculate a + b. Comparing $(60)^2 S(60) = 3599(61)^{61} + 61$ with the given form $a(b)^b + b$, we can identify: $a = 3599$ $b = 61$.

Both $a$ and $b$ are natural numbers ($N$). Now, calculate $a+b$: $a+b = 3599 + 61 = 3660$.

Common Mistakes & Tips

• Sign Errors: Be extremely careful when subtracting series and manipulating equations. A small sign error can lead to an incorrect final answer.
• Geometric Series Formula Application: Ensure the correct first term, common ratio, and number of terms are used when applying the geometric series sum formula.
• Substitution Order: Substitute the value of $x$ only after obtaining the simplified closed-form expression for $S(x)$ to avoid complex calculations.

Summary

The given series $\mathrm{S}(x)$ is an arithmetico-geometric series. By multiplying the series by its common ratio $(1+x)$ and subtracting it from the original series, we were able to transform it into a geometric series, allowing us to derive a closed-form expression for $S(x)$. Substituting $x=60$ into this expression and comparing it with the given form $a(b)^b + b$, we found $a=3599$ and $b=61$. The sum $a+b$ is $3660$.

The final answer is \boxed{3660}.