JEE Main 2019 (12 January)

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Animated Solution for Mathematics - Sequence and Series: 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

Analyze the Given Series

Given series: $(\frac{3}{4})^{3} + (1 \frac{1}{2})^{3} + (2 \frac{1}{4})^{3} + 3^{3} + \dots$
Number of terms to sum: $n = 15$
Total Sum: $S_{15} = 225 k$

Convert to Improper Fractions

Let's rewrite the mixed fractions to spot a pattern.
$T_{1} = (\frac{3}{4})^{3}$
$T_{2} = (1 \frac{1}{2})^{3} = (\frac{3}{2})^{3} = (\frac{6}{4})^{3}$
$T_{3} = (2 \frac{1}{4})^{3} = (\frac{9}{4})^{3}$
$T_{4} = 3^{3} = (\frac{12}{4})^{3}$

Find the General Term $T_{r}$

The numerators are multiples of $3$ : $3, 6, 9, 12, \dots$
The denominator is constant: $4$
General term: $T_{r} = (\frac{3 r}{4})^{3}$

Express the Sum in Sigma Notation

The sum of the first $15$ terms is:
$S_{15} = \sum_{r = 1}^{15} T_{r}$
$S_{15} = \sum_{r = 1}^{15} (\frac{3 r}{4})^{3}$

Factor Out the Constants

Expand the cube: $(\frac{3 r}{4})^{3} = \frac{27 r ^{3}}{64}$
Factor out the constant term from the summation:
$S_{15} = \frac{27}{64} \sum_{r = 1}^{15} r^{3}$

Apply the Sum of Cubes Formula

Standard Identity for the sum of first $n$ cubes:
$\sum_{r = 1}^{n} r^{3} = [\frac{n ( n + 1 )}{2}]^{2}$
For our problem, the upper limit is $n = 15$ .

Substitute $n = 15$ and Simplify

Substitute $n = 15$ into the formula:
$\sum_{r = 1}^{15} r^{3} = [\frac{15 ( 15 + 1 )}{2}]^{2}$
$\sum_{r = 1}^{15} r^{3} = [\frac{15 \times 16}{2}]^{2}$
$\sum_{r = 1}^{15} r^{3} = (15 \times 8)^{2}$

Calculate the Sum of Cubes

Square the result:
$\sum_{r = 1}^{15} r^{3} = 12 0^{2}$
$12 0^{2} = 14400$

Calculate the Total Sum $S_{15}$

Substitute the summation value back into $S_{15}$ :
$S_{15} = \frac{27}{64} \times 14400$
Divide $14400$ by $64$ :
$\frac{14400}{64} = 225$
$S_{15} = 27 \times 225$

Solve for $k$

Equate the calculated sum to the given form:
$225 k = 27 \times 225$
Divide both sides by $225$ :
$k = 27$

Conclusion and Key Takeaway

Final Answer: $k = 27$
Key Takeaway: Identifying the general term $T_{r}$ is the foundation for solving complex series problems.
Standard Formulas: Always remember $\sum r$ , $\sum r^{2}$ , and $\sum r^{3}$ for JEE.

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The Sigma Insight: Sum of Special Series

Analyzing the Setup

The given series is:

(\frac{3}{4})^{3} + (1 \frac{1}{2})^{3} + (2 \frac{1}{4})^{3} + 3^{3} + \dots

To solve this, we must identify the underlying pattern. We are tasked with finding the sum of the first 15 terms, denoted as

S_{15} = 225 k

Normalizing the Chaos

First, we convert all terms into improper fractions to reveal the structure:

T_{1} = (\frac{3}{4})^{3}

T_{2} = (1 \frac{1}{2})^{3} = (\frac{3}{2})^{3} = (\frac{6}{4})^{3}

T_{3} = (2 \frac{1}{4})^{3} = (\frac{9}{4})^{3}

T_{4} = 3^{3} = (\frac{12}{4})^{3}

The numerators follow the arithmetic progression

3, 6, 9, 12, \dots

, which can be written as

3 r

for the

r

-th term. The denominator remains a constant

4

. Thus, the general term is:

T_{r} = (\frac{3 r}{4})^{3}

The Power of Sigma

Now that we have the general term

T_{r} = \frac{27 r ^{3}}{64}

, we express the sum of the first 15 terms using Sigma notation:

S_{15} = r = 1 \sum 15 (\frac{3 r}{4})^{3}

By expanding the cube and pulling the constant factor out of the summation, we get:

S_{15} = \frac{27}{64} r = 1 \sum 15 r^{3}

The Elegant Conclusion

We utilize the standard identity for the sum of cubes of the first

n

natural numbers:

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

For

n = 15

, the calculation is:

[\frac{15 ( 16 )}{2}]^{2} = (15 \times 8)^{2} = 12 0^{2} = 14400

Substituting this back into our expression for

S_{15}

S_{15} = \frac{27}{64} \times 14400

Since

14400 \div 64 = 225

, the equation simplifies to:

S_{15} = 27 \times 225

Given that

S_{15} = 225 k

, we equate the two:

225 k = 27 \times 225

By cancelling

225

from both sides, we find the final value: $k = 27$

Common Traps & FAQs

Why did we convert to improper fractions? Can't we just work with mixed fractions?

Working with mixed fractions is like trying to run a race with your shoelaces tied together. By converting to improper fractions, we reveal the underlying arithmetic progression in the numerators, which is the key to unlocking the entire problem.

Why is the denominator constant at 4? Is that always the case?

In this specific series, yes. The denominator is a constant because the terms were constructed that way. Always look for what stays the same and what changes. The constant denominator is a gift—it allows us to factor it out of the summation entirely.

Is there a shortcut for calculating

12 0^{2}

Absolutely! Whenever you square a number ending in zero, square the non-zero part and append double the number of zeros.

1 2^{2} = 144

, so

12 0^{2} = 14400

. It's a small trick, but it saves precious time in the exam hall.

Animated Solution for Mathematics - Sequence and Series: 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

Analyze the Given Series

Convert to Improper Fractions

Find the General Term $T_{r}$

Express the Sum in Sigma Notation

Factor Out the Constants

Apply the Sum of Cubes Formula

Substitute $n = 15$ and Simplify

Calculate the Sum of Cubes

Calculate the Total Sum $S_{15}$

Solve for $k$

Conclusion and Key Takeaway

The Sigma Insight: Sum of Special Series

Analyzing the Setup

Normalizing the Chaos

The Power of Sigma

The Elegant Conclusion

Similar Questions

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

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

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 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 $\frac{1}{1 - 3 \cdot 1 ^{2} + 1 ^{4}} + \frac{2}{1 - 3 \cdot 2 ^{2} + 2 ^{4}} + \frac{3}{1 - 3 \cdot 3 ^{2} + 3 ^{4}} + \dots$ up to 10 terms is

The sum of $10$ 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 :

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

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 Given Series

Convert to Improper Fractions

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

Express the Sum in Sigma Notation

Factor Out the Constants

Apply the Sum of Cubes Formula

Substitute .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } n=15 and Simplify

Calculate the Sum of Cubes

Calculate the Total Sum .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } S15​

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

Conclusion and Key Takeaway

The Sigma Insight: Sum of Special Series

Analyzing the Setup

Normalizing the Chaos

The Power of Sigma

The Elegant Conclusion

Similar Questions

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Find the General Term $T_{r}$

Substitute $n = 15$ and Simplify

Calculate the Total Sum $S_{15}$

Solve for $k$

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

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

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 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 $\frac{1}{1 - 3 \cdot 1 ^{2} + 1 ^{4}} + \frac{2}{1 - 3 \cdot 2 ^{2} + 2 ^{4}} + \frac{3}{1 - 3 \cdot 3 ^{2} + 3 ^{4}} + \dots$ up to 10 terms is

The sum of $10$ 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 :

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

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