If $\left(\frac{1}{\alpha+1} + \frac{1}{\alpha+2} + \dots + \frac{1}{\alpha+1012}\right) - \left(\frac{1}{2 \cdot 1} + \frac{1}{4 \cdot 3} + \frac{1}{6 \cdot 5} + \dots + \frac{1}{2024 \cdot 2023}\right) = \frac{1}{2024}$, then $\alpha$ is equal to

### Analyzing the Setup We are given two series, $S_1$ and $S_2$, defined as follows: $S_1 = \sum_{k=1}^{1012} \frac{1}{\alpha+k}$ $S_2 = \sum_{k=1}^{1012} \frac{1}{2k(2k-1)}$ Our objective is to determine the value of $\alpha$ given the constraint $S_1 - S_2 = \frac{1}{2024}$. ### Deconstructing the Second Series The general term of $S_2$ can be decomposed using partial fractions: $$\frac{1}{2k(2k-1)} = \frac{1}{2k-1} - \frac{1}{2k}$$ Summing this from $k=1$ to $1012$, we obtain an alternating harmonic series: $$S_2 = \left( 1 - \frac{1}{2} \right) + \left( \frac{1}{3} - \frac{1}{4} \right) + \dots + \left( \frac{1}{2023} - \frac{1}{2024} \right)$$ To simplify this, we employ the "add and subtract" technique. We add the even terms $\left( \frac{1}{2} + \frac{1}{4} + \dots + \frac{1}{2024} \right)$ to the series and subtract them twice to maintain equality: $$S_2 = \left( 1 + \frac{1}{2} + \frac{1}{3} + \dots + \frac{1}{2024} \right) - 2 \left( \frac{1}{2} + \frac{1}{4} + \dots + \frac{1}{2024} \right)$$ ### Simplifying the Expression The second part of the expression simplifies significantly: $$2 \left( \frac{1}{2} + \frac{1}{4} + \dots + \frac{1}{2024} \right) = 1 + \frac{1}{2} + \dots + \frac{1}{1012}$$ Substituting this back, $S_2$ becomes the difference between the harmonic sum up to $2024$ and the harmonic sum up to $1012$: $$S_2 = \sum_{n=1}^{2024} \frac{1}{n} - \sum_{n=1}^{1012} \frac{1}{n} = \sum_{n=1013}^{2024} \frac{1}{n}$$ ### Solving for Alpha Now, we substitute $S_2$ into the original equation $S_1 = S_2 + \frac{1}{2024}$: $$S_1 = \left( \frac{1}{1013} + \frac{1}{1014} + \dots + \frac{1}{2024} \right) + \frac{1}{2024}$$ Combining the last two terms, we get $\frac{1}{2024} + \frac{1}{2024} = \frac{2}{2024} = \frac{1}{1012}$. Thus: $$S_1 = \frac{1}{1012} + \frac{1}{1013} + \dots + \frac{1}{2023}$$ Comparing this to the definition $S_1 = \sum_{k=1}^{1012} \frac{1}{\alpha+k}$, we equate the first terms: $$\alpha + 1 = 1012$$ Therefore, the final value is: **$\alpha = 1011$**

JEE Main 2024 (09 Apr Shift 2)

LEVELJEE Main

Animated Solution for Mathematics - Sequence and Series: If $(\frac{1}{α + 1} + \frac{1}{α + 2} + \dots + \frac{1}{α + 1012}) - (\frac{1}{2 \cdot 1} + \frac{1}{4 \cdot 3} + \frac{1}{6 \cdot 5} + \dots + \frac{1}{2024 \cdot 2023}) = \frac{1}{2024}$ , then $α$ is equal to

Q: How did the alternating series turn into a sum of reciprocals? I am confused by the 'add and subtract' trick.

Think of it as balancing an equation. We have $1 - \frac{1}{2} + \frac{1}{3} - \frac{1}{4} + \dots$. By adding the even terms, we complete the harmonic series $1 + \frac{1}{2} + \frac{1}{3} + \frac{1}{4} + \dots$. But since we added them, we must subtract them twice to keep the value unchanged. It is a classic technique to transform alternating series into standard ones.

Enter Numerical Value:

Visualized Solution

Defining the Series $S_{1}$ and $S_{2}$

Let $S_{1} = \sum_{k = 1}^{1012} \frac{1}{α + k}$
Let $S_{2} = \sum_{k = 1}^{1012} \frac{1}{2 k ( 2 k - 1 )}$
The given equation is $S_{1} - S_{2} = \frac{1}{2024}$

Decomposing $S_{2}$ using Partial Fractions

General term of $S_{2}$ : $T_{k} = \frac{1}{2 k ( 2 k - 1 )}$
Using partial fractions: $\frac{1}{2 k ( 2 k - 1 )} = \frac{1}{2 k - 1} - \frac{1}{2 k}$
So, $S_{2} = \sum_{k = 1}^{1012} (\frac{1}{2 k - 1} - \frac{1}{2 k})$

Expanding the Alternating Series

Expand $S_{2}$ for $k = 1, 2, \dots, 1012$ :
$S_{2} = (1 - \frac{1}{2}) + (\frac{1}{3} - \frac{1}{4}) + \dots + (\frac{1}{2023} - \frac{1}{2024})$
$S_{2} = 1 - \frac{1}{2} + \frac{1}{3} - \frac{1}{4} + \dots + \frac{1}{2023} - \frac{1}{2024}$

The 'Add and Subtract' Trick

Add and subtract even terms: $(\frac{1}{2} + \frac{1}{4} + \dots + \frac{1}{2024})$
$S_{2} = (1 + \frac{1}{2} + \frac{1}{3} + \dots + \frac{1}{2024}) - 2 (\frac{1}{2} + \frac{1}{4} + \dots + \frac{1}{2024})$

Simplifying the Subtracted Sum

Simplify the second part: $2 (\frac{1}{2} + \frac{1}{4} + \dots + \frac{1}{2024}) = (1 + \frac{1}{2} + \dots + \frac{1}{1012})$
$S_{2} = \sum_{k = 1}^{2024} \frac{1}{k} - \sum_{k = 1}^{1012} \frac{1}{k}$
$S_{2} = \sum_{k = 1013}^{2024} \frac{1}{k}$

Substituting $S_{2}$ back into the Equation

Given: $S_{1} - S_{2} = \frac{1}{2024} \Rightarrow S_{1} = S_{2} + \frac{1}{2024}$
Substitute $S_{2}$ : $S_{1} = (\frac{1}{1013} + \frac{1}{1014} + \dots + \frac{1}{2024}) + \frac{1}{2024}$

Handling the Extra Term

$S_{1} = \frac{1}{1013} + \frac{1}{1014} + \dots + \frac{1}{2023} + (\frac{1}{2024} + \frac{1}{2024})$
$S_{1} = \frac{1}{1013} + \frac{1}{1014} + \dots + \frac{1}{2023} + \frac{2}{2024}$
$S_{1} = \frac{1}{1013} + \frac{1}{1014} + \dots + \frac{1}{2023} + \frac{1}{1012}$

Matching the Structure of $S_{1}$

Rearrange in order: $S_{1} = \frac{1}{1012} + \frac{1}{1013} + \dots + \frac{1}{2023}$
Original form: $S_{1} = \frac{1}{α + 1} + \frac{1}{α + 2} + \dots + \frac{1}{α + 1012}$

Solving for $α$

Compare the first terms: $α + 1 = 1012$
$α = 1011$
Check the last terms: $α + 1012 = 1011 + 1012 = 2023$ (Matches!)

Final Conclusion

Key Takeaway: The alternating series $1 - \frac{1}{2} + \frac{1}{3} - \dots - \frac{1}{2 n}$ is always equal to $\sum_{k = n + 1}^{2 n} \frac{1}{k}$ .
Final Answer: $α = 1011$

00:00 / 00:00

The Sigma Insight: Sum of Special Series

Analyzing the Setup

We are given two series,

S_{1}

and

S_{2}

, defined as follows:

S_{1} = \sum_{k = 1}^{1012} \frac{1}{α + k}

S_{2} = \sum_{k = 1}^{1012} \frac{1}{2 k ( 2 k - 1 )}

Our objective is to determine the value of

α

given the constraint

S_{1} - S_{2} = \frac{1}{2024}

Deconstructing the Second Series

The general term of

S_{2}

can be decomposed using partial fractions:

\frac{1}{2 k ( 2 k - 1 )} = \frac{1}{2 k - 1} - \frac{1}{2 k}

Summing this from

k = 1

1012

, we obtain an alternating harmonic series:

S_{2} = (1 - \frac{1}{2}) + (\frac{1}{3} - \frac{1}{4}) + \dots + (\frac{1}{2023} - \frac{1}{2024})

To simplify this, we employ the "add and subtract" technique. We add the even terms

(\frac{1}{2} + \frac{1}{4} + \dots + \frac{1}{2024})

to the series and subtract them twice to maintain equality:

S_{2} = (1 + \frac{1}{2} + \frac{1}{3} + \dots + \frac{1}{2024}) - 2 (\frac{1}{2} + \frac{1}{4} + \dots + \frac{1}{2024})

Simplifying the Expression

The second part of the expression simplifies significantly:

2 (\frac{1}{2} + \frac{1}{4} + \dots + \frac{1}{2024}) = 1 + \frac{1}{2} + \dots + \frac{1}{1012}

Substituting this back,

S_{2}

becomes the difference between the harmonic sum up to

2024

and the harmonic sum up to

1012

S_{2} = n = 1 \sum 2024 \frac{1}{n} - n = 1 \sum 1012 \frac{1}{n} = n = 1013 \sum 2024 \frac{1}{n}

Solving for Alpha

Now, we substitute

S_{2}

into the original equation

S_{1} = S_{2} + \frac{1}{2024}

S_{1} = (\frac{1}{1013} + \frac{1}{1014} + \dots + \frac{1}{2024}) + \frac{1}{2024}

Combining the last two terms, we get

\frac{1}{2024} + \frac{1}{2024} = \frac{2}{2024} = \frac{1}{1012}

. Thus:

S_{1} = \frac{1}{1012} + \frac{1}{1013} + \dots + \frac{1}{2023}

Comparing this to the definition

S_{1} = \sum_{k = 1}^{1012} \frac{1}{α + k}

, we equate the first terms:

α + 1 = 1012

Therefore, the final value is: $α = 1011$

Common Traps & FAQs

Why can't we just calculate the sum directly? It looks like a simple arithmetic series.

Ah, the trap! It is not an arithmetic series. It is a harmonic series. You cannot sum these directly because the denominators are changing in a non-linear way. We must use algebraic manipulation to find the hidden pattern.

How did the alternating series turn into a sum of reciprocals? I am confused by the 'add and subtract' trick.

Think of it as balancing an equation. We have

1 - \frac{1}{2} + \frac{1}{3} - \frac{1}{4} + \dots

. By adding the even terms, we complete the harmonic series

1 + \frac{1}{2} + \frac{1}{3} + \frac{1}{4} + \dots

. But since we added them, we must subtract them twice to keep the value unchanged. It is a classic technique to transform alternating series into standard ones.

Why does

α + 1 = 1012

? How do we know the terms match?

We have two series of the same length. If the sums are equal and the terms are in the same harmonic progression, the individual terms must correspond. By aligning the first term of the first series with the first term of the second, we lock the value of

α

Animated Solution for Mathematics - Sequence and Series: If $(\frac{1}{α + 1} + \frac{1}{α + 2} + \dots + \frac{1}{α + 1012}) - (\frac{1}{2 \cdot 1} + \frac{1}{4 \cdot 3} + \frac{1}{6 \cdot 5} + \dots + \frac{1}{2024 \cdot 2023}) = \frac{1}{2024}$ , then $α$ is equal to

Enter Numerical Value:

Visualized Solution

Defining the Series $S_{1}$ and $S_{2}$

Decomposing $S_{2}$ using Partial Fractions

Expanding the Alternating Series

The 'Add and Subtract' Trick

Simplifying the Subtracted Sum

Substituting $S_{2}$ back into the Equation

Handling the Extra Term

Matching the Structure of $S_{1}$

Solving for $α$

Final Conclusion

The Sigma Insight: Sum of Special Series

Analyzing the Setup

Deconstructing the Second Series

Simplifying the Expression

Solving for Alpha

Similar Questions

If $\frac{1}{1 ^{4}} + \frac{1}{2 ^{4}} + \frac{1}{3 ^{4}} + \dots \infty = \frac{π ^{4}}{90}$ , $\frac{1}{1 ^{4}} + \frac{1}{3 ^{4}} + \frac{1}{5 ^{4}} + \dots \infty = α$ , $\frac{1}{2 ^{4}} + \frac{1}{4 ^{4}} + \frac{1}{6 ^{4}} + \dots \infty = β$ , then $\frac{α}{β}$ is equal to

If $\sum_{r = 1}^{10} r! (r^{3} + 6 r^{2} + 2 r + 5) = α (11!)$ , then the value of $α$ is equal to

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Let $S_{k} = \frac{1 + 2 + 3 + ... + k}{k}$ . If $S_{1}^{2} + S_{2}^{2} + ... + S_{10}^{2} = \frac{5}{12} A$ , then A is equal to :

For $k \in N$ , if the sum of the series $1 + \frac{4}{k} + \frac{8}{k ^{2}} + \frac{13}{k ^{3}} + \frac{19}{k ^{4}} + \dots$ is 10, then the value of $k$ is

$\frac{2 ^{3} - 1 ^{3}}{1 \times 7} + \frac{4 ^{3} - 3 ^{3} + 2 ^{3} - 1 ^{3}}{2 \times 11} + \frac{6 ^{3} - 5 ^{3} + 4 ^{3} - 3 ^{3} + 2 ^{3} - 1 ^{3}}{3 \times 15} + \dots + \frac{3 0 ^{3} - 2 9 ^{3} + 2 8 ^{3} - 2 7 ^{3} + \dots + 2 ^{3} - 1 ^{3}}{15 \times 63}$ is equal to ________.

Enter Numerical Value:

Visualized Solution

Defining the Series .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } S1​ and .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } S2​

Decomposing .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } S2​ using Partial Fractions

Expanding the Alternating Series

The 'Add and Subtract' Trick

Simplifying the Subtracted Sum

Substituting .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } S2​ back into the Equation

Handling the Extra Term

Matching the Structure of .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } S1​

Solving for .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } α

Final Conclusion

The Sigma Insight: Sum of Special Series

Analyzing the Setup

Deconstructing the Second Series

Simplifying the Expression

Solving for Alpha

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Defining the Series $S_{1}$ and $S_{2}$

Decomposing $S_{2}$ using Partial Fractions

Substituting $S_{2}$ back into the Equation

Matching the Structure of $S_{1}$

Solving for $α$

If $\frac{1}{1 ^{4}} + \frac{1}{2 ^{4}} + \frac{1}{3 ^{4}} + \dots \infty = \frac{π ^{4}}{90}$ , $\frac{1}{1 ^{4}} + \frac{1}{3 ^{4}} + \frac{1}{5 ^{4}} + \dots \infty = α$ , $\frac{1}{2 ^{4}} + \frac{1}{4 ^{4}} + \frac{1}{6 ^{4}} + \dots \infty = β$ , then $\frac{α}{β}$ is equal to

If $\sum_{r = 1}^{10} r! (r^{3} + 6 r^{2} + 2 r + 5) = α (11!)$ , then the value of $α$ is equal to

$\frac{1}{3 ^{2} - 1} + \frac{1}{5 ^{2} - 1} + \frac{1}{7 ^{2} - 1} + \dots + \frac{1}{( 201 ) ^{2} - 1}$ is equal to

$\frac{6}{3 ^{26}} + \frac{10 \cdot 1}{3 ^{25}} + \frac{10 \cdot 2}{3 ^{24}} + \frac{10 \cdot 2 ^{2}}{3 ^{23}} + \dots + \frac{10 \cdot 2 ^{24}}{3}$ is equal to :

Let $S_{k} = \frac{1 + 2 + 3 + ... + k}{k}$ . If $S_{1}^{2} + S_{2}^{2} + ... + S_{10}^{2} = \frac{5}{12} A$ , then A is equal to :

For $k \in N$ , if the sum of the series $1 + \frac{4}{k} + \frac{8}{k ^{2}} + \frac{13}{k ^{3}} + \frac{19}{k ^{4}} + \dots$ is 10, then the value of $k$ is

$\frac{2 ^{3} - 1 ^{3}}{1 \times 7} + \frac{4 ^{3} - 3 ^{3} + 2 ^{3} - 1 ^{3}}{2 \times 11} + \frac{6 ^{3} - 5 ^{3} + 4 ^{3} - 3 ^{3} + 2 ^{3} - 1 ^{3}}{3 \times 15} + \dots + \frac{3 0 ^{3} - 2 9 ^{3} + 2 8 ^{3} - 2 7 ^{3} + \dots + 2 ^{3} - 1 ^{3}}{15 \times 63}$ is equal to ________.