JEE Main 2012

LEVELJEE Main

Animated Solution for Mathematics - Sequence and Series: Statement-1: The sum of the series $1 + (1 + 2 + 4) + (4 + 6 + 9) + (9 + 12 + 16) + \dots + (361 + 380 + 400)$ is $8000$ . Statement-2: $\sum_{k = 1}^{n} (k^{3} - (k - 1)^{3}) = n^{3}$ , for any natural number $n$ .

Q: Wait, why is the first term just 1? The pattern seems to start from the second bracket.

Great observation! The series is $1 + (1+2+4) + (4+6+9) + \dots$. If we look at the general term $T_k = (k-1)^2 + (k-1)k + k^2$, for $k=1$, we get $0^2 + 0(1) + 1^2 = 1$. So, the first term $1$ actually fits the pattern perfectly as $T_1$!

Q: How do I know the series ends at $k=20$?

Look at the last bracket: $(361 + 380 + 400)$. Notice $361 = 19^2$ and $400 = 20^2$. This matches $(k-1)^2$ and $k^2$ when $k=20$. Since the pattern is $(k-1)^2 + (k-1)k + k^2$, the last term is $T_{20}$.

Select Answer:

Visualized Solution

Analyzing the Problem Structure

We are given two statements to evaluate.
Statement-1 claims the sum of a specific series is $8000$ .
Statement-2 presents a general identity: $\sum_{k = 1}^{n} (k^{3} - (k - 1)^{3}) = n^{3}$ .
Let's first analyze Statement-2 to see if it can serve as our primary tool.

Expanding the Telescoping Sum

Let's expand the summation in Statement-2: $\sum_{k = 1}^{n} (k^{3} - (k - 1)^{3})$
For $k = 1$ : $1^{3} - 0^{3}$
For $k = 2$ : $2^{3} - 1^{3}$
For $k = 3$ : $3^{3} - 2^{3}$
This structure is a classic example of a telescoping series.

Visualizing the Cancellation

Sum $= (1^{3} - 0^{3}) + (2^{3} - 1^{3}) + (3^{3} - 2^{3}) + \dots + (n^{3} - (n - 1)^{3})$
Notice how $+ 1^{3}$ cancels with $- 1^{3}$ .
Similarly, $+ 2^{3}$ cancels with $- 2^{3}$ .
This chain reaction continues all the way to the end.

The Surviving Terms

Only two terms survive this massive cancellation:
The very first negative term: $- 0^{3}$
The very last positive term: $+ n^{3}$
Therefore, the sum simplifies to: $n^{3} - 0^{3} = n^{3}$
Statement-2 is True.

Analyzing Statement-1

Let's write down the series in Statement-1:
$S = 1 + (1 + 2 + 4) + (4 + 6 + 9) + (9 + 12 + 16) + \dots + (361 + 380 + 400)$
We need to find a general formula for the $k$ -th term, $T_{k}$ .
Let's examine the structure of each bracketed term.

Decoding $T_{1}$ and $T_{2}$

First term: $T_{1} = 1$
Second term: $T_{2} = 1 + 2 + 4$
Notice that $1 = 1^{2}$ , $4 = 2^{2}$ , and $2 = 1 \cdot 2$ .
So, $T_{2} = 1^{2} + 1 \cdot 2 + 2^{2}$ .
This looks like the quadratic part of the cubic difference identity!

Decoding $T_{3}$

Third term: $T_{3} = 4 + 6 + 9$
Notice that $4 = 2^{2}$ , $9 = 3^{2}$ , and $6 = 2 \cdot 3$ .
So, $T_{3} = 2^{2} + 2 \cdot 3 + 3^{2}$ .
The pattern is now extremely clear!

The General Term $T_{k}$

For the $k$ -th term, the numbers involved are $(k - 1)$ and $k$ .
General term: $T_{k} = (k - 1)^{2} + (k - 1) k + k^{2}$
Recall the identity: $a^{3} - b^{3} = (a - b) (a^{2} + ab + b^{2})$
If we set $a = k$ and $b = k - 1$ , then $(a - b) = 1$ .
Therefore, $T_{k} = k^{3} - (k - 1)^{3}$ .

Finding the Number of Terms

The last term is: $361 + 380 + 400$
Notice that $361 = 1 9^{2}$ , $400 = 2 0^{2}$ , and $380 = 19 \cdot 20$ .
This matches our general term $T_{k}$ for $k = 20$ .
So, there are exactly $20$ terms in this series.

Applying the Telescoping Tool

The sum of the series is: $S = \sum_{k = 1}^{20} T_{k}$
Substituting $T_{k}$ : $S = \sum_{k = 1}^{20} (k^{3} - (k - 1)^{3})$
Using Statement-2 with $n = 20$ :
$S = 2 0^{3}$

Final Evaluation

Calculate $2 0^{3}$ :
$2 0^{3} = 20 \cdot 20 \cdot 20 = 8000$
This matches the value in Statement-1.
Statement-1 is True.
Since we used Statement-2 to prove Statement-1, Statement-2 is the correct explanation.

00:00 / 00:00

The Sigma Insight: Sum of Special Series

Analyzing the Telescoping Identity

Statement-2 provides the identity:

k = 1 \sum n (k^{3} - (k - 1)^{3}) = n^{3}

This is a classic example of a telescoping series. When we expand the summation, we observe the following terms: For

k = 1

1^{3} - 0^{3}

For

k = 2

2^{3} - 1^{3}

For

k = 3

3^{3} - 2^{3}

As the series progresses, the positive term of one bracket cancels the negative term of the subsequent bracket. This chain reaction continues until only the very first negative term,

- 0^{3}

, and the very last positive term,

n^{3}

, remain. Thus, the sum simplifies to

n^{3}

, confirming that Statement-2 is true.

Deconstructing the Main Series

Now, let us evaluate the sum

S

given in Statement-1:

S = 1 + (1 + 2 + 4) + (4 + 6 + 9) + (9 + 12 + 16) + \dots + (361 + 380 + 400)

By observing the structure of each bracket, we identify a clear pattern. The second term is

1^{2} + 1 \cdot 2 + 2^{2}

, and the third term is

2^{2} + 2 \cdot 3 + 3^{2}

. Generally, the

k

-th term

T_{k}

can be expressed as:

T_{k} = (k - 1)^{2} + (k - 1) k + k^{2}

Applying the Algebraic Identity

Recall the algebraic identity for the difference of two cubes:

a^{3} - b^{3} = (a - b) (a^{2} + ab + b^{2})

If we set

a = k

and

b = k - 1

, then

(a - b) = 1

. Consequently, the expression for

T_{k}

simplifies perfectly:

T_{k} = k^{3} - (k - 1)^{3}

Our series

S

is the sum of these terms from

k = 1

k = 20

. Note that the first term

1

can be written as

1^{3} - 0^{3}

Final Calculation

Using the telescoping property proven in Statement-2, the sum of the series is:

S = k = 1 \sum 20 (k^{3} - (k - 1)^{3}) = 2 0^{3}

Calculating the final value:

2 0^{3} = 8000

The logic is consistent and the derivation is sound. Therefore, Statement-1 is true, and Statement-2 is the correct explanation for the result.

Common Traps & FAQs

Wait, why is the first term just 1? The pattern seems to start from the second bracket.

Great observation! The series is

1 + (1 + 2 + 4) + (4 + 6 + 9) + \dots

. If we look at the general term

T_{k} = (k - 1)^{2} + (k - 1) k + k^{2}

, for

k = 1

, we get

0^{2} + 0 (1) + 1^{2} = 1

. So, the first term

1

actually fits the pattern perfectly as

T_{1}

How do I know the series ends at

k = 20

Look at the last bracket:

(361 + 380 + 400)

. Notice

361 = 1 9^{2}

and

400 = 2 0^{2}

. This matches

(k - 1)^{2}

and

k^{2}

when

k = 20

. Since the pattern is

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

, the last term is

T_{20}

Do I really need Statement-2 to solve Statement-1?

You could calculate the sum manually, but that would be tedious and prone to error. Statement-2 provides the 'telescoping' property, which is the elegant shortcut. In JEE, recognizing these patterns is the difference between finishing in time and getting stuck!

Animated Solution for Mathematics - Sequence and Series: Statement-1: The sum of the series $1 + (1 + 2 + 4) + (4 + 6 + 9) + (9 + 12 + 16) + \dots + (361 + 380 + 400)$ is $8000$ . Statement-2: $\sum_{k = 1}^{n} (k^{3} - (k - 1)^{3}) = n^{3}$ , for any natural number $n$ .

Select Answer:

Visualized Solution

Analyzing the Problem Structure

Expanding the Telescoping Sum

Visualizing the Cancellation

The Surviving Terms

Analyzing Statement-1

Decoding $T_{1}$ and $T_{2}$

Decoding $T_{3}$

The General Term $T_{k}$

Finding the Number of Terms

Applying the Telescoping Tool

Final Evaluation

The Sigma Insight: Sum of Special Series

Analyzing the Telescoping Identity

Deconstructing the Main Series

Applying the Algebraic Identity

Final Calculation

Similar Questions

Sum of the first $n$ terms of the series $\frac{1}{2} + \frac{3}{4} + \frac{7}{8} + \frac{15}{16} + \dots$ is equal to

Statement-1 : For every natural number $n \geq 2, \frac{1}{1} + \frac{1}{2} + \dots + \frac{1}{n} > n$ . Statement-2 : For every natural number $n \geq 2, n (n + 1) < n + 1$ .

The sum of first $9$ terms of the series $\frac{1 ^{3}}{1} + \frac{1 ^{3} + 2 ^{3}}{1 + 3} + \frac{1 ^{3} + 2 ^{3} + 3 ^{3}}{1 + 3 + 5} + \dots$ 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

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The sum of the first $n$ terms of the series $1^{2} + 2. 2^{2} + 3^{2} + 2. 4^{2} + 5^{2} + 2. 6^{2} + \dots$ is $n (n + 1)^{2} /2$ when $n$ is even. When $n$ is odd the sum is

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Let $S (K) = 1 + 3 + 5 + ... + (2 K - 1) = 3 + K^{2}$ . Then which of the following is true

If the sum of the first 15 terms of the series $(3/4)^{3} + (1/2)^{3} + (2/3)^{3} + 3^{3} + (3/4)^{3} + \dots$ is equal to $225 k$ , then $k$ is equal to :

Select Answer:

Visualized Solution

Analyzing the Problem Structure

Expanding the Telescoping Sum

Visualizing the Cancellation

The Surviving Terms

Analyzing Statement-1

Decoding .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } T1​ and .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } T2​

Decoding .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } T3​

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

Finding the Number of Terms

Applying the Telescoping Tool

Final Evaluation

The Sigma Insight: Sum of Special Series

Analyzing the Telescoping Identity

Deconstructing the Main Series

Applying the Algebraic Identity

Final Calculation

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Decoding $T_{3}$

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