Question

If \[ S(x) = (1 + x) + 2(1 + x)^2 + 3(1 + x)^3 + \ldots + 60(1 + x)^{60}, \, x \neq 0, \] and \[ (60)^2 S(60) = a(b)^b + b, \] where $a, b \in \mathbb{N}$, then $(a + b)$ is equal to ________.

Answer 1

Correct Answer: 3660

Starting with the series:
\[S(x) = (1 + x) + 2(1 + x)^2 + 3(1 + x)^3 + \dots + 60(1 + x)^{60}\]
Multiplying both sides by \( (1 + x) \), we get:
\[(1 + x)S = (1 + x) + 2(1 + x)^2 + 3(1 + x)^3 + \dots + 60(1 + x)^{61}\]
Now, subtracting \( S \) from \( (1 + x)S \), we obtain:
\[-xS = \frac{(1 + x)(1 + x)^{60} - 1}{x} - 60(1 + x)^{61}\]
Now, put \( x = 60 \):
\[-60S = \frac{61((61)^{60} - 1)}{60} - 60 \cdot (61)^{61}\]
Solving this equation gives:
\[S = 3660\]

Answer 2

The problem asks to find the sum of a given Arithmetico-Geometric Progression (AGP), \(S(x)\). We then need to evaluate this sum for \(x=60\) and use the given relation \((60)^2 S(60) = a(b)^b + b\) to find the values of natural numbers \(a\) and \(b\). Finally, we are asked to compute the value of \((a+b)\).

Concept Used:

The given series is an Arithmetico-Geometric Progression (AGP). An AGP is a series where each term is the product of a term from an arithmetic progression and a term from a geometric progression. To find the sum of a finite AGP, a standard method is used:

1. Let the sum be \(S\).
2. Multiply the series by the common ratio \(r\) of the geometric part.
3. Subtract the new series (\(rS\)) from the original series (\(S\)). This subtraction results in a new series that is primarily a geometric progression, which can be summed easily using the formula for the sum of a finite GP.

Step-by-Step Solution:

Step 1: Express the series \(S(x)\) as an AGP.

Let \(y = 1+x\). The series \(S(x)\) can be rewritten as:

\[ S(x) = y + 2y^2 + 3y^3 + \ldots + 60y^{60} \quad \text{(Equation 1)} \]

This is an AGP with an arithmetic part \(1, 2, 3, \ldots, 60\) and a geometric part \(y, y^2, y^3, \ldots, y^{60}\) with a common ratio of \(y\).

Step 2: Apply the method for summing an AGP.

Multiply Equation 1 by the common ratio \(y\):

\[ yS(x) = y^2 + 2y^3 + 3y^4 + \ldots + 59y^{60} + 60y^{61} \quad \text{(Equation 2)} \]

Subtract Equation 2 from Equation 1:

\[ S(x) - yS(x) = (y + 2y^2 + \ldots + 60y^{60}) - (y^2 + 2y^3 + \ldots + 60y^{61}) \] \[ (1-y)S(x) = y + (2y^2 - y^2) + (3y^3 - 2y^3) + \ldots + (60y^{60} - 59y^{60}) - 60y^{61} \] \[ (1-y)S(x) = (y + y^2 + y^3 + \ldots + y^{60}) - 60y^{61} \]

The series in the parenthesis is a finite GP with 60 terms, first term \(y\), and common ratio \(y\). Its sum is \(\frac{y(y^{60}-1)}{y-1}\).

\[ (1-y)S(x) = \frac{y(y^{60}-1)}{y-1} - 60y^{61} \]

Step 3: Substitute \(x = 60\) into the derived expression.

For \(x = 60\), we have \(y = 1+60 = 61\). Consequently, \(1-y = -60\) and \(y-1 = 60\).

Substitute these values into the sum formula:

\[ (-60)S(60) = \frac{61(61^{60}-1)}{61-1} - 60(61)^{61} \] \[ -60 S(60) = \frac{61(61^{60}-1)}{60} - 60(61)^{61} \]

Step 4: Rearrange the equation to find an expression for \((60)^2 S(60)\).

Multiply the entire equation by \(-60\):

\[ (-60)(-60)S(60) = -60 \left( \frac{61(61^{60}-1)}{60} \right) - (-60) \left( 60(61)^{61} \right) \] \[ (60)^2 S(60) = -61(61^{60}-1) + (60)^2(61)^{61} \] \[ (60)^2 S(60) = -(61^{61}-61) + 3600(61)^{61} \] \[ (60)^2 S(60) = -61^{61} + 61 + 3600(61)^{61} \] \[ (60)^2 S(60) = (3600 - 1)61^{61} + 61 \] \[ (60)^2 S(60) = 3599(61)^{61} + 61 \]

Step 5: Compare with the given format to find \(a\) and \(b\).

We are given the relation:

\[ (60)^2 S(60) = a(b)^b + b \]

Comparing this with our result:

\[ 3599(61)^{61} + 61 = a(b)^b + b \]

By direct comparison, we can identify:

\[ a = 3599 \quad \text{and} \quad b = 61 \]

These are both natural numbers as required.

Step 6: Calculate the final value of \((a+b)\).

The final step is to sum the values of \(a\) and \(b\):

\[ a + b = 3599 + 61 \] \[ a + b = 3660 \]

Thus, the value of \((a+b)\) is 3660.