JEE Main 2025 April

LEVELJEE Main

Animated Solution for Mathematics - Sequence and Series: If the sum of the first 10 terms of the series $\frac{4.1}{1 + 4. 1 ^{4}} + \frac{4.2}{1 + 4. 2 ^{4}} + \frac{4.3}{1 + 4. 3 ^{4}} + \dots$ is $\frac{m}{n}$ , where $g cd (m, n) = 1$ , then $m + n$ is equal to______

Q: How do I know for sure that this is a telescoping series?

A telescoping series is one where the general term $T_r$ can be written as $f(r) - f(r+1)$ (or some variation). When you sum these up, the $-f(r+1)$ of one term cancels with the $+f(r+1)$ of the next. In our case, if you let $f(r) = \frac{1}{2r^2-2r+1}$, you will find that $f(r+1) = \frac{1}{2(r+1)^2-2(r+1)+1} = \frac{1}{2r^2+2r+1}$. Since $T_r = f(r) - f(r+1)$, the cancellation is guaranteed!

Enter Numerical Value:

Visualized Solution

Identifying the General Term $T_{r}$

Observe the pattern of the given series: $\frac{4 \cdot 1}{1 + 4 \cdot 1 ^{4}} + \frac{4 \cdot 2}{1 + 4 \cdot 2 ^{4}} + \dots$
The $r^{t h}$ term of the series can be written as:
$T_{r} = \frac{4 r}{1 + 4 r ^{4}}$

Factorizing the Denominator $1 + 4 r^{4}$

To factorize the denominator $1 + 4 r^{4}$ , we use the technique of completing the square.
Add and subtract $4 r^{2}$ :
$1 + 4 r^{4} = (1 + 4 r^{4} + 4 r^{2}) - 4 r^{2}$
This is an application of the Sophie Germain Identity.

Applying Difference of Squares

Rewrite the expression as a difference of squares:
$1 + 4 r^{4} = (1 + 2 r^{2})^{2} - (2 r)^{2}$
Using the identity $a^{2} - b^{2} = (a - b) (a + b)$ :
$1 + 4 r^{4} = (1 + 2 r^{2} - 2 r) (1 + 2 r^{2} + 2 r)$

Partial Fraction Decomposition

Notice the difference between the two factors in the denominator:
$(2 r^{2} + 2 r + 1) - (2 r^{2} - 2 r + 1) = 4 r$
Substitute this difference into the numerator of $T_{r}$ :
$T_{r} = \frac{( 2 r ^{2} + 2 r + 1 ) - ( 2 r ^{2} - 2 r + 1 )}{( 2 r ^{2} - 2 r + 1 ) ( 2 r ^{2} + 2 r + 1 )}$
$T_{r} = \frac{1}{2 r ^{2} - 2 r + 1} - \frac{1}{2 r ^{2} + 2 r + 1}$

Setting up the Telescoping Sum

The sum of the first 10 terms is $S_{10} = \sum_{r = 1}^{10} T_{r}$ .
Let's evaluate the first few terms:
For $r = 1 : T_{1} = \frac{1}{1} - \frac{1}{5}$
For $r = 2 : T_{2} = \frac{1}{5} - \frac{1}{13}$
For $r = 3 : T_{3} = \frac{1}{13} - \frac{1}{25}$
The pattern of cancellation becomes clear.

Calculating the Last Term

We continue this process up to the $1 0^{t h}$ term.
For $r = 10 : T_{10} = \frac{1}{2 ( 10 ) ^{2} - 2 ( 10 ) + 1} - \frac{1}{2 ( 10 ) ^{2} + 2 ( 10 ) + 1}$
$T_{10} = \frac{1}{200 - 20 + 1} - \frac{1}{200 + 20 + 1}$
$T_{10} = \frac{1}{181} - \frac{1}{221}$

Final Result for $m / n$

Summing all terms: $S_{10} = (1 - \frac{1}{5}) + (\frac{1}{5} - \frac{1}{13}) + \dots + (\frac{1}{181} - \frac{1}{221})$
After massive cancellation, only the first and last terms remain:
$S_{10} = 1 - \frac{1}{221}$
$S_{10} = \frac{221 - 1}{221} = \frac{220}{221}$
Comparing with $\frac{m}{n}$ , we get $m = 220$ and $n = 221$ .

Final Answer: $m + n$

The question asks for the value of $m + n$ .
We have $m = 220$ and $n = 221$ .
Check if $g cd (m, n) = 1$ : $g cd (220, 221) = 1$ . (Condition satisfied)
Substitute the values:
$m + n = 220 + 221 = 441$
Final Answer: 441

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The Sigma Insight: Sum of Special Series

Analyzing the Setup

Imagine you are standing before a complex, intimidating series. You see

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

and your first instinct might be to panic.

But as an elite JEE aspirant, you know that every complex problem is just a simple one in disguise. Let us embark on a journey to unravel this.

The General Term

First, we must identify the heartbeat of the series: the general term

T_{r}

. By observing the pattern, it is clear that the

r^{t h}

term is:

T_{r} = \frac{4 r}{1 + 4 r ^{4}}

This denominator,

1 + 4 r^{4}

, is the villain of our story. It looks stubborn, but it has a secret weakness.

The Sophie Germain Identity

We need to break down

1 + 4 r^{4}

. We use the technique of completing the square by treating

1

1^{2}

and

4 r^{4}

(2 r^{2})^{2}

To make this a perfect square, we need a middle term of

2 \cdot 1 \cdot 2 r^{2} = 4 r^{2}

. So, we add and subtract

4 r^{2}

1 + 4 r^{4} = (1 + 4 r^{4} + 4 r^{2}) - 4 r^{2}

This transforms our expression into

(1 + 2 r^{2})^{2} - (2 r)^{2}

. Now, we have a difference of squares,

a^{2} - b^{2} = (a - b) (a + b)

, which gives us:

(2 r^{2} - 2 r + 1) (2 r^{2} + 2 r + 1)

This is the Sophie Germain Identity in action, and it has completely dismantled our villain.

The Bridge to Telescoping

Now, look at the factors:

(2 r^{2} - 2 r + 1)

and

(2 r^{2} + 2 r + 1)

. If we subtract the smaller from the larger, we find:

(2 r^{2} + 2 r + 1) - (2 r^{2} - 2 r + 1) = 4 r

This is exactly our numerator! This allows us to write

T_{r}

as:

T_{r} = \frac{( 2 r ^{2} + 2 r + 1 ) - ( 2 r ^{2} - 2 r + 1 )}{( 2 r ^{2} - 2 r + 1 ) ( 2 r ^{2} + 2 r + 1 )}

This simplifies to the elegant form:

T_{r} = \frac{1}{2 r ^{2} - 2 r + 1} - \frac{1}{2 r ^{2} + 2 r + 1}

The Domino Effect

Now, the magic happens. We are summing from

r = 1

r = 10

. Let

f (r) = \frac{1}{2 r ^{2} - 2 r + 1}

. Then

T_{r} = f (r) - f (r + 1)

When we write out the sum

S_{10} = \sum_{r = 1}^{10} (f (r) - f (r + 1))

, we get:

(f (1) - f (2)) + (f (2) - f (3)) + \dots + (f (10) - f (11))

Everything in the middle cancels out, leaving us with

S_{10} = f (1) - f (11)

Final Calculation

Substituting

r = 1

, we find

f (1) = \frac{1}{2 ( 1 ) ^{2} - 2 ( 1 ) + 1} = 1

Substituting

r = 10

into the second part, we find

f (11) = \frac{1}{2 ( 10 ) ^{2} + 2 ( 10 ) + 1} = \frac{1}{221}

Thus, the sum is:

S_{10} = 1 - \frac{1}{221} = \frac{220}{221}

We have

m = 220

and

n = 221

. Since

g cd (220, 221) = 1

, the condition is satisfied. The final answer is $m + n = 220 + 221 = 441$ . You have conquered the series!

Common Traps & FAQs

Wait, why do we add and subtract

4 r^{2}

in the denominator? How would I ever think of that?

That is a fantastic question! This is a classic technique called 'completing the square' to force a difference of squares. In competitive exams like JEE Advanced, whenever you see a denominator of the form

a^{4} + 4 b^{4}

, it is a huge hint to use the Sophie Germain Identity. It is a pattern recognition skill that you build with practice. Once you see

1 + 4 r^{4}

, think of it as

1^{4} + 4 (r)^{4}

and try to complete the square.

How do I know for sure that this is a telescoping series?

A telescoping series is one where the general term

T_{r}

can be written as

f (r) - f (r + 1)

(or some variation). When you sum these up, the

- f (r + 1)

of one term cancels with the

+ f (r + 1)

of the next. In our case, if you let

f (r) = \frac{1}{2 r ^{2} - 2 r + 1}

, you will find that

f (r + 1) = \frac{1}{2 ( r + 1 ) ^{2} - 2 ( r + 1 ) + 1} = \frac{1}{2 r ^{2} + 2 r + 1}

. Since

T_{r} = f (r) - f (r + 1)

, the cancellation is guaranteed!

Why did the middle terms cancel out so perfectly?

It is the beauty of the telescoping structure! When you write out the sum

S_{10} = (f (1) - f (2)) + (f (2) - f (3)) + \dots + (f (10) - f (11))

, you see that

- f (2)

cancels with

+ f (2)

- f (3)

with

+ f (3)

, and so on. Only the very first part,

f (1)

, and the very last part,

- f (11)

, survive the massacre. It is like a row of dominoes falling; everything in the middle is knocked down, leaving only the ends standing.

Animated Solution for Mathematics - Sequence and Series: If the sum of the first 10 terms of the series $\frac{4.1}{1 + 4. 1 ^{4}} + \frac{4.2}{1 + 4. 2 ^{4}} + \frac{4.3}{1 + 4. 3 ^{4}} + \dots$ is $\frac{m}{n}$ , where $g cd (m, n) = 1$ , then $m + n$ is equal to______

Enter Numerical Value:

Visualized Solution

Identifying the General Term $T_{r}$

Factorizing the Denominator $1 + 4 r^{4}$

Applying Difference of Squares

Partial Fraction Decomposition

Setting up the Telescoping Sum

Calculating the Last Term

Final Result for $m / n$

Final Answer: $m + n$

The Sigma Insight: Sum of Special Series

Analyzing the Setup

The General Term

The Sophie Germain Identity

The Bridge to Telescoping

The Domino Effect

Final Calculation

Similar Questions

If the sum of the first 20 terms of the series $\frac{4.1}{4 + 3. 1 ^{2} + 1 ^{4}} + \frac{4.2}{4 + 3. 2 ^{2} + 2 ^{4}} + \frac{4.3}{4 + 3. 3 ^{2} + 3 ^{4}} + \frac{4.4}{4 + 3. 4 ^{2} + 4 ^{4}} + \dots$ is $\frac{m}{n}$ , where $m$ and $n$ are coprime, then $m + n$ is equal to :-

If the sum of the first ten terms of the series $\frac{1}{5} + \frac{2}{65} + \frac{3}{325} + \frac{4}{1025} + \frac{5}{2501} + \dots$ is $\frac{m}{n}$ , where $m$ and $n$ are co-prime numbers, then $m + n$ is equal to ______.

The sum to 10 terms of the series $\frac{1}{1 + 1 ^{2} + 1 ^{4}} + \frac{2}{1 + 2 ^{2} + 2 ^{4}} + \frac{3}{1 + 3 ^{2} + 3 ^{4}} + \dots$ is:-

If $\sum_{k = 1}^{10} \frac{k}{k ^{4} + k ^{2} + 1} = \frac{m}{n}$ , where $m$ and $n$ are co-prime, then $m + n$ is equal to ________.

If the sum of the first ten terms of the series $(1 \frac{3}{5})^{2} + (2 \frac{2}{5})^{2} + (3 \frac{1}{5})^{2} + 4^{2} + (4 \frac{4}{5})^{2} + \dots$ is $\frac{16}{5} m$ , then $m$ is equal to:

The sum of the series $\frac{1}{1 - 3 \cdot 1 ^{2} + 1 ^{4}} + \frac{2}{1 - 3 \cdot 2 ^{2} + 2 ^{4}} + \frac{3}{1 - 3 \cdot 3 ^{2} + 3 ^{4}} + \dots$ up to 10 terms is

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

The sum of $10$ terms of the series $\frac{3}{1 ^{2} \times 2 ^{2}} + \frac{5}{2 ^{2} \times 3 ^{2}} + \frac{7}{3 ^{2} \times 4 ^{2}} + \dots$ is :

The sum of the series $1 + 2 \times 3 + 3 \times 5 + 4 \times 7 + \dots$ upto 11th term is :-

Enter Numerical Value:

Visualized Solution

Identifying the General Term .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Tr​

Factorizing the Denominator .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } 1+4r4

Applying Difference of Squares

Partial Fraction Decomposition

Setting up the Telescoping Sum

Calculating the Last Term

Final Result for .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } m/n

Final Answer: .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } m+n

The Sigma Insight: Sum of Special Series

Analyzing the Setup

The General Term

The Sophie Germain Identity

The Bridge to Telescoping

The Domino Effect

Final Calculation

Similar Questions

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Identifying the General Term $T_{r}$

Factorizing the Denominator $1 + 4 r^{4}$

Final Result for $m / n$

Final Answer: $m + n$

If the sum of the first ten terms of the series $\frac{1}{5} + \frac{2}{65} + \frac{3}{325} + \frac{4}{1025} + \frac{5}{2501} + \dots$ is $\frac{m}{n}$ , where $m$ and $n$ are co-prime numbers, then $m + n$ is equal to ______.

The sum to 10 terms of the series $\frac{1}{1 + 1 ^{2} + 1 ^{4}} + \frac{2}{1 + 2 ^{2} + 2 ^{4}} + \frac{3}{1 + 3 ^{2} + 3 ^{4}} + \dots$ is:-

If $\sum_{k = 1}^{10} \frac{k}{k ^{4} + k ^{2} + 1} = \frac{m}{n}$ , where $m$ and $n$ are co-prime, then $m + n$ is equal to ________.

If the sum of the first ten terms of the series $(1 \frac{3}{5})^{2} + (2 \frac{2}{5})^{2} + (3 \frac{1}{5})^{2} + 4^{2} + (4 \frac{4}{5})^{2} + \dots$ is $\frac{16}{5} m$ , then $m$ is equal to:

The sum of the series $\frac{1}{1 - 3 \cdot 1 ^{2} + 1 ^{4}} + \frac{2}{1 - 3 \cdot 2 ^{2} + 2 ^{4}} + \frac{3}{1 - 3 \cdot 3 ^{2} + 3 ^{4}} + \dots$ up to 10 terms is

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

The sum of $10$ terms of the series $\frac{3}{1 ^{2} \times 2 ^{2}} + \frac{5}{2 ^{2} \times 3 ^{2}} + \frac{7}{3 ^{2} \times 4 ^{2}} + \dots$ is :

The sum of the series $1 + 2 \times 3 + 3 \times 5 + 4 \times 7 + \dots$ upto 11th term is :-