Why did we re-index the sum from $1$ to $10$ instead of just calculating $2^2 + 3^2 + \dots + 10^2$ directly?

It's about leveraging the standard formula $\sum_{k=1}^{m} k^2$. We know the formula for $1$ to $m$. It is much faster to calculate the sum from $1$ to $10$ and subtract the $1^2$ term than to manually sum nine squares. Efficiency is the name of the game!

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Animated Solution for Mathematics - Sequence and Series: 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

Q: Wait, why is the denominator $n^2$? I thought it was just an arithmetic progression.

It is! The denominator is $1 + 3 + 5 + \dots + (2n-1)$. This is an AP with first term $a=1$ and common difference $d=2$. The sum is $\frac{n}{2}[2(1) + (n-1)2] = \frac{n}{2}[2 + 2n - 2] = n^2$. Geometrically, it's even cooler: visualize a square of side $n$ built by adding layers of odd numbers!

Select Answer:

Visualized Solution

Analyze the Series Pattern

Observe the given series: $\frac{1 ^{3}}{1} + \frac{1 ^{3} + 2 ^{3}}{1 + 3} + \frac{1 ^{3} + 2 ^{3} + 3 ^{3}}{1 + 3 + 5} + \dots$
Let's break down each term into its numerator and denominator.
The $n^{t h}$ term, $T_{n}$ , will have a numerator of $1^{3} + 2^{3} + \dots + n^{3}$ and a denominator of $1 + 3 + \dots + (2 n - 1)$ .

Formulate the General Term $T_{n}$

The general term is: $T_{n} = \frac{1 ^{3} + 2 ^{3} + 3 ^{3} + \dots + n ^{3}}{1 + 3 + 5 + \dots + ( 2 n - 1 )}$
We can write this compactly using summation notation:
$T_{n} = \frac{\sum _{k = 1}^{n} k ^{3}}{\sum _{k = 1}^{n} ( 2 k - 1 )}$

Simplify the Numerator

Recall the standard formula for the sum of the first $n$ cubes:
$\sum_{k = 1}^{n} k^{3} = [\frac{n ( n + 1 )}{2}]^{2}$
This simplifies to: $\frac{n ^{2} ( n + 1 ) ^{2}}{4}$

Simplify the Denominator

Recall the formula for the sum of the first $n$ odd numbers:
$\sum_{k = 1}^{n} (2 k - 1) = n^{2}$
We can also visualize this geometrically: adding consecutive odd numbers always forms a perfect square!

Combine and Simplify $T_{n}$

Substitute the simplified numerator and denominator back into $T_{n}$ :
$T_{n} = \frac{\frac{n ^{2} ( n + 1 ) ^{2}}{4}}{n ^{2}}$
Cancel the common term $n^{2}$ from the numerator and denominator:
$T_{n} = \frac{( n + 1 ) ^{2}}{4}$

Set up the Summation $S_{9}$

We need to find the sum of the first $9$ terms: $S_{9} = \sum_{n = 1}^{9} T_{n}$
Substitute the simplified $T_{n}$ :
$S_{9} = \sum_{n = 1}^{9} \frac{( n + 1 ) ^{2}}{4}$
Factor out the constant $\frac{1}{4}$ :
$S_{9} = \frac{1}{4} \sum_{n = 1}^{9} (n + 1)^{2}$

Expand and Re-index the Sum

Let's write out the terms of the sum:
$\sum_{n = 1}^{9} (n + 1)^{2} = 2^{2} + 3^{2} + 4^{2} + \dots + 1 0^{2}$
To use standard formulas, we can rewrite this as:
$(\sum_{k = 1}^{10} k^{2}) - 1^{2}$

Calculate the Sum of Squares

Recall the formula for the sum of the first $m$ squares:
$\sum_{k = 1}^{m} k^{2} = \frac{m ( m + 1 ) ( 2 m + 1 )}{6}$
For $m = 10$ :
$\sum_{k = 1}^{10} k^{2} = \frac{10 \times 11 \times 21}{6} = 385$
Subtracting the missing $1^{2}$ term: $385 - 1 = 384$

Final Calculation and Result

Substitute the sum back into the expression for $S_{9}$ :
$S_{9} = \frac{1}{4} \times 384$
Divide $384$ by $4$ :
$S_{9} = 96$
The sum of the first $9$ terms is 96 (Option 4).

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The Sigma Insight: Sum of Special Series

Solution Diagram

Analyzing the Setup

The series is given by:

\frac{1 ^{3}}{1} + \frac{1 ^{3} + 2 ^{3}}{1 + 3} + \frac{1 ^{3} + 2 ^{3} + 3 ^{3}}{1 + 3 + 5} + \dots

To solve this, we must identify the general term,

T_{n}

. The numerator is the sum of the first

n

cubes, and the denominator is the sum of the first

n

odd numbers.

We express this as:

T_{n} = \frac{\sum _{k = 1}^{n} k ^{3}}{\sum _{k = 1}^{n} ( 2 k - 1 )}

The Algebraic Alchemy

We apply standard mathematical identities to simplify the expression. The numerator is a well-known identity:

k = 1 \sum n k^{3} = [\frac{n ( n + 1 )}{2}]^{2} = \frac{n ^{2} ( n + 1 ) ^{2}}{4}

The denominator represents the sum of the first

n

odd numbers, which simplifies elegantly to

n^{2}

. Substituting these back into our expression for

T_{n}

, we observe a significant simplification:

T_{n} = \frac{\frac{n ^{2} ( n + 1 ) ^{2}}{4}}{n ^{2}}

The

n^{2}

terms cancel out perfectly, leaving us with the simplified general term:

T_{n} = \frac{( n + 1 ) ^{2}}{4}

The Grand Summation

We are tasked with finding the sum of the first

9

terms,

S_{9} = \sum_{n = 1}^{9} T_{n}

. Substituting our simplified

T_{n}

, we obtain:

S_{9} = \frac{1}{4} n = 1 \sum 9 (n + 1)^{2}

Expanding this summation, we get

2^{2} + 3^{2} + \dots + 1 0^{2}

. This is equivalent to the sum of the first

10

squares, excluding the

1^{2}

term:

S_{9} = \frac{1}{4} [(k = 1 \sum 10 k^{2}) - 1^{2}]

Using the standard formula

\sum_{k = 1}^{m} k^{2} = \frac{m ( m + 1 ) ( 2 m + 1 )}{6}

for

m = 10

\frac{10 \times 11 \times 21}{6} = 385

Subtracting the

1^{2}

term yields

384

. Finally, we calculate the total sum:

S_{9} = \frac{1}{4} \times 384 = 96

The final result is 96.

Common Traps & FAQs

Wait, why is the denominator

n^{2}

? I thought it was just an arithmetic progression.

It is! The denominator is

1 + 3 + 5 + \dots + (2 n - 1)

. This is an AP with first term

a = 1

and common difference

d = 2

. The sum is

\frac{n}{2} [2 (1) + (n - 1) 2] = \frac{n}{2} [2 + 2 n - 2] = n^{2}

. Geometrically, it's even cooler: visualize a square of side

n

built by adding layers of odd numbers!

Why can't I just calculate the first 9 terms manually?

You absolutely could, and in an exam, if you're stuck, do it! But the JEE Advanced is about finding the *structure*. By deriving the general term, you ensure accuracy and speed, and you build the mathematical maturity needed for harder problems where

n

might be

100

1000

Why did we re-index the sum from

1

10

instead of just calculating

2^{2} + 3^{2} + \dots + 1 0^{2}

directly?

It's about leveraging the standard formula

\sum_{k = 1}^{m} k^{2}

. We know the formula for

1

m

. It is much faster to calculate the sum from

1

10

and subtract the

1^{2}

term than to manually sum nine squares. Efficiency is the name of the game!

Animated Solution for Mathematics - Sequence and Series: 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

Select Answer:

Visualized Solution

Analyze the Series Pattern

Formulate the General Term $T_{n}$

Simplify the Numerator

Simplify the Denominator

Combine and Simplify $T_{n}$

Set up the Summation $S_{9}$

Expand and Re-index the Sum

Calculate the Sum of Squares

Final Calculation and Result

The Sigma Insight: Sum of Special Series

Analyzing the Setup

The Algebraic Alchemy

The Grand Summation

Similar Questions

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 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:-

The sum of the first 20 terms of the series $1 + \frac{3}{2} + \frac{7}{4} + \frac{15}{8} + \frac{31}{16} + \dots$ is :

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

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

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

The sum of the following series $1 + 6 + \frac{9 ( 1 ^{2} + 2 ^{2} + 3 ^{2} )}{7} + \frac{12 ( 1 ^{2} + 2 ^{2} + 3 ^{2} + 4 ^{2} )}{9} + \frac{15 ( 1 ^{2} + 2 ^{2} + ... + 5 ^{2} )}{11} + ...$ up to 15 terms, is:

The sum of the series $1 + \frac{1}{4.2 !} + \frac{1}{16.4 !} + \frac{1}{64.6 !} + \dots$ ad inf. is

The sum of the first 20 terms of the series $5 + 11 + 19 + 29 + 41 + \dots$ is

The sum $\frac{3 \cdot 1 ^{3}}{1 ^{2}} + \frac{5 \cdot ( 1 ^{3} + 2 ^{3} )}{1 ^{2} + 2 ^{2}} + \frac{7 \cdot ( 1 ^{3} + 2 ^{3} + 3 ^{3} )}{1 ^{2} + 2 ^{2} + 3 ^{2}} + \dots$ is :

Animated Solution for Mathematics - Sequence and Series: .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } The sum of first 9 terms of the series 113​+1+313+23​+1+3+513+23+33​+… is

Select Answer:

Visualized Solution

Analyze the Series Pattern

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

Simplify the Numerator

Simplify the Denominator

Combine and Simplify .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Tn​

Set up the Summation .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } S9​

Expand and Re-index the Sum

Calculate the Sum of Squares

Final Calculation and Result

The Sigma Insight: Sum of Special Series

Analyzing the Setup

The Algebraic Alchemy

The Grand Summation

Similar Questions

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Animated Solution for Mathematics - Sequence and Series: 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

Formulate the General Term $T_{n}$

Combine and Simplify $T_{n}$

Set up the Summation $S_{9}$

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 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:-

The sum of the first 20 terms of the series $1 + \frac{3}{2} + \frac{7}{4} + \frac{15}{8} + \frac{31}{16} + \dots$ is :

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

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

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

The sum of the following series $1 + 6 + \frac{9 ( 1 ^{2} + 2 ^{2} + 3 ^{2} )}{7} + \frac{12 ( 1 ^{2} + 2 ^{2} + 3 ^{2} + 4 ^{2} )}{9} + \frac{15 ( 1 ^{2} + 2 ^{2} + ... + 5 ^{2} )}{11} + ...$ up to 15 terms, is:

The sum of the series $1 + \frac{1}{4.2 !} + \frac{1}{16.4 !} + \frac{1}{64.6 !} + \dots$ ad inf. is

The sum of the first 20 terms of the series $5 + 11 + 19 + 29 + 41 + \dots$ is

The sum $\frac{3 \cdot 1 ^{3}}{1 ^{2}} + \frac{5 \cdot ( 1 ^{3} + 2 ^{3} )}{1 ^{2} + 2 ^{2}} + \frac{7 \cdot ( 1 ^{3} + 2 ^{3} + 3 ^{3} )}{1 ^{2} + 2 ^{2} + 3 ^{2}} + \dots$ is :