For k ∈ N, let 1/(α(α+1)(α+2)...(α+20)) = Σ(k=0 to 20) Ak/(α+k), where α 0. Then the value of 100(

Q: Wait, why does the cover-up rule work? Isn't it just a trick?

It is not a trick, but a fundamental property of partial fraction decomposition! By multiplying both sides by $(\alpha+k)$, you effectively remove the singularity at $\alpha = -k$. When you then take the limit as $\alpha \to -k$, all other terms on the right side still contain a factor of $(\alpha+k)$ in their numerator, so they vanish. Only the $A_k$ term remains, giving you its exact value. It is a powerful surgical tool in algebra.

Q: How do I handle the negative signs in the factorial product?

That is a great question! When you substitute $\alpha = -k$, the product of terms before the missing factor is $(-k)(-k+1)\dots(-1)$. There are exactly $k$ such terms. If you factor out a $(-1)$ from each of these $k$ terms, you get $(-1)^k$. What remains is $k imes (k-1) imes \dots imes 1$, which is simply $k!$. So, the product becomes $(-1)^k k!$. It is all about keeping track of the signs!

Q: Is there a faster way to calculate the ratio $\frac{A_{14} + A_{15}}{A_{13}}$ without calculating the full factorials?

Yes, absolutely! Instead of computing the massive factorial values, write out the ratios $\frac{A_{14}}{A_{13}}$ and $\frac{A_{15}}{A_{13}}$ algebraically. You will find that most of the factorials cancel out beautifully, leaving you with simple fractions. This is the beauty of these problems—the complexity often collapses into something elegant.

Question

For $k \in \mathbb{N}$, let $\frac{1}{\alpha(\alpha+1)(\alpha+2)\dots(\alpha+20)} = \sum_{k=0}^{20} \frac{A_k}{\alpha+k}$, where $\alpha > 0$. Then the value of $100 \left( \frac{A_{14} + A_{15}}{A_{13}} \right)^2$ is equal to

Sigma Percentile · Accepted Answer

### Analyzing the Setup

We are presented with the partial fraction decomposition of a product of twenty-one terms:

$$\frac{1}{\alpha(\alpha+1)(\alpha+2)\dots(\alpha+20)} = \sum_{k=0}^{20} \frac{A_k}{\alpha+k}$$

Our goal is to determine the coefficients $A_k$ and evaluate the expression $100 \left( \frac{A_{14} + A_{15}}{A_{13}} ight)^2$.

### The Surgical Strike (The Cover-up Rule)

To isolate a specific coefficient $A_k$, we employ the **Cover-up Rule**. We multiply both sides of the equation by $(\alpha+k)$ and take the limit as $\alpha 	o -k$.

This causes all terms on the right side, except for $A_k$, to vanish. The coefficient is defined as:

$$A_k = \lim_{\alpha 	o -k} \frac{\alpha+k}{\alpha(\alpha+1)\dots(\alpha+k)\dots(\alpha+20)}$$

### The Factorial Dance

When we substitute $\alpha = -k$, the denominator splits into two products. The terms before the missing factor are $(-k), (-k+1), \dots, (-1)$, which simplifies to $(-1)^k \cdot k!$.

The terms after the missing factor are $1, 2, \dots, (20-k)$, which simplifies to $(20-k)!$. Combining these, we obtain the general formula for the coefficients:

$$A_k = \frac{1}{(-1)^k \cdot k! \cdot (20-k)!}$$

### Calculating the Coefficients

Using our formula, we calculate the specific values required for the final expression:

For $k=13$:
$$A_{13} = \frac{1}{(-1)^{13} \cdot 13! \cdot 7!} = -\frac{1}{13! \cdot 7!}$$

For $k=14$:
$$A_{14} = \frac{1}{(-1)^{14} \cdot 14! \cdot 6!} = \frac{1}{14! \cdot 6!}$$

For $k=15$:
$$A_{15} = \frac{1}{(-1)^{15} \cdot 15! \cdot 5!} = -\frac{1}{15! \cdot 5!}$$

### Final Calculation

First, we compute the ratios relative to $A_{13}$:

$$\frac{A_{14}}{A_{13}} = \frac{1}{14! \cdot 6!} \cdot \left( -\frac{13! \cdot 7!}{1} ight) = -\frac{7}{14} = -\frac{1}{2}$$

$$\frac{A_{15}}{A_{13}} = \left( -\frac{1}{15! \cdot 5!} ight) \cdot \left( -\frac{13! \cdot 7!}{1} ight) = \frac{6 \cdot 7}{14 \cdot 15} = \frac{42}{210} = \frac{1}{5}$$

Summing these ratios gives $-\frac{1}{2} + \frac{1}{5} = -\frac{3}{10}$. Squaring this result and multiplying by $100$ yields:

$$100 \left( -\frac{3}{10} ight)^2 = 100 \cdot \frac{9}{100} = 9$$

The final answer is **9**.

Animated Solution for Mathematics - Indefinite Integration: For $k \in N$ , let $\frac{1}{α ( α + 1 ) ( α + 2 ) \dots ( α + 20 )} = \sum_{k = 0}^{20} \frac{A _{k}}{α + k}$ , where $α > 0$ . Then the value of $100 (\frac{A _{14} + A _{15}}{A _{13}})^{2}$ is equal to

Enter Numerical Value:

Visualized Solution

The Partial Fraction Setup

The Cover-up Rule for $A_{k}$

Evaluating the Product for $A_{k}$

Expressing $A_{k}$ using Factorials

Calculating $A_{13}$

Calculating $A_{14}$

Calculating $A_{15}$

Ratio $\frac{A _{14}}{A _{13}}$

Ratio $\frac{A _{15}}{A _{13}}$

Summing the Ratios

Final Calculation

The Sigma Insight: Integration using Partial Fractions

Analyzing the Setup

The Surgical Strike (The Cover-up Rule)

The Factorial Dance

Calculating the Coefficients

Final Calculation

Similar Questions

If $\int \frac{s i n x}{s i n ^{3} x + c o s ^{3} x} d x = α lo g_{e} ∣1 + tan x ∣ + β lo g_{e} ∣1 - tan x + tan^{2} x ∣ + γ tan^{- 1} (\frac{2 t a n x - 1}{3}) + C$ , when $C$ is constant of integration, then the value of $18 (α + β + γ^{2})$ is.

Integrate $\int \frac{x ^{3} + 3 x + 2}{( x ^{2} + 1 ) ^{2} ( x + 1 )} d x$

Enter Numerical Value:

Visualized Solution

The Partial Fraction Setup

The Cover-up Rule for .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Ak​

Evaluating the Product for .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Ak​

Expressing .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Ak​ using Factorials

Calculating .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } A13​

Calculating .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } A14​

Calculating .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } A15​

Ratio .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } A13​A14​​

Ratio .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } A13​A15​​

Summing the Ratios

Final Calculation

The Sigma Insight: Integration using Partial Fractions

Analyzing the Setup

The Surgical Strike (The Cover-up Rule)

The Factorial Dance

Calculating the Coefficients

Final Calculation

Similar Questions

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Integrate ∫(x2+1)2(x+1)x3+3x+2​dx

The Cover-up Rule for $A_{k}$

Evaluating the Product for $A_{k}$

Expressing $A_{k}$ using Factorials

Calculating $A_{13}$

Calculating $A_{14}$

Calculating $A_{15}$

Ratio $\frac{A _{14}}{A _{13}}$

Ratio $\frac{A _{15}}{A _{13}}$

Integrate $\int \frac{x ^{3} + 3 x + 2}{( x ^{2} + 1 ) ^{2} ( x + 1 )} d x$