JEE Main 2024 (06 Apr Shift 1)

LEVELJEE Main

Animated Solution for Mathematics - Sequence and Series: Let the first term of a series be $T_{1} = 6$ and its $r^{th}$ term $T_{r} = 3 T_{r - 1} + 6^{r}, r = 2, 3, \dots, n$ . If the sum of the first $n$ terms of this series is $\frac{1}{5} (n^{2} - 12 n + 39) (4 \cdot 6^{n} - 5 \cdot 3^{n} + 1)$ , then $n$ is equal to

Q: Can we solve this using mathematical induction?

While you could verify the answer for small $n$ using induction, it is not the most efficient way to derive the expression. The algebraic method of normalization and summation is more robust and provides deeper insight into the structure of the series.

Q: Why does the summation for $b_n$ start from $r=2$?

The recurrence $T_r = 3T_{r-1} + 6^r$ is defined for $r = 2, 3, \dots, n$. Therefore, the difference $b_r - b_{r-1} = 2^r$ is also valid for $r \geq 2$. We start from $b_1$ and add the differences from $r=2$ up to $n$.

Enter Numerical Value:

Visualized Solution

The Recurrence Relation

Given recurrence: $T_{r} = 3 T_{r - 1} + 6^{r}$ for $r = 2, 3, \dots, n$
Divide both sides by $3^{r}$ :
$\frac{T _{r}}{3 ^{r}} = \frac{3 T _{r - 1}}{3 ^{r}} + \frac{6 ^{r}}{3 ^{r}}$
Simplify the terms:
$\frac{T _{r}}{3 ^{r}} = \frac{T _{r - 1}}{3 ^{r - 1}} + 2^{r}$

Defining a New Sequence $b_{r}$

Let $b_{r} = \frac{T _{r}}{3 ^{r}}$
The relation becomes: $b_{r} = b_{r - 1} + 2^{r}$
Rearranging gives the difference: $b_{r} - b_{r - 1} = 2^{r}$

Calculating $b_{n}$ using Summation

$b_{n} = b_{1} + \sum_{r = 2}^{n} 2^{r}$
Given $T_{1} = 6$ , so $b_{1} = \frac{T _{1}}{3 ^{1}} = \frac{6}{3} = 2$
$b_{n} = 2 + (2^{2} + 2^{3} + \dots + 2^{n})$
The terms in parentheses form a G.P. with $a = 4$ , $r = 2$ , and $n - 1$ terms.

Evaluating the G.P. Sum

Sum of G.P. $= \frac{4 ( 2 ^{n - 1} - 1 )}{2 - 1} = 4 \cdot 2^{n - 1} - 4 = 2^{n + 1} - 4$
Substitute back into $b_{n}$ :
$b_{n} = 2 + (2^{n + 1} - 4)$
$b_{n} = 2^{n + 1} - 2$

Finding the General Term $T_{n}$

$T_{n} = 3^{n} \cdot b_{n}$
$T_{n} = 3^{n} (2^{n + 1} - 2)$
$T_{n} = 3^{n} (2 \cdot 2^{n} - 2)$
$T_{n} = 2 \cdot (3 \cdot 2)^{n} - 2 \cdot 3^{n}$
$T_{n} = 2 \cdot 6^{n} - 2 \cdot 3^{n}$

Setting up the Sum $S_{n}$

$S_{n} = \sum_{r = 1}^{n} T_{r} = \sum_{r = 1}^{n} (2 \cdot 6^{r} - 2 \cdot 3^{r})$
$S_{n} = 2 \sum_{r = 1}^{n} 6^{r} - 2 \sum_{r = 1}^{n} 3^{r}$

Summing the Individual G.P.s

Sum 1: $\sum_{r = 1}^{n} 6^{r} = \frac{6 ( 6 ^{n} - 1 )}{6 - 1} = \frac{6}{5} (6^{n} - 1)$
Sum 2: $\sum_{r = 1}^{n} 3^{r} = \frac{3 ( 3 ^{n} - 1 )}{3 - 1} = \frac{3}{2} (3^{n} - 1)$
Substitute into $S_{n}$ :
$S_{n} = 2 [\frac{6}{5} (6^{n} - 1)] - 2 [\frac{3}{2} (3^{n} - 1)]$
$S_{n} = \frac{12}{5} (6^{n} - 1) - 3 (3^{n} - 1)$

Combining and Simplifying $S_{n}$

$S_{n} = \frac{12 ( 6 ^{n} - 1 ) - 15 ( 3 ^{n} - 1 )}{5}$
$S_{n} = \frac{12 \cdot 6 ^{n} - 12 - 15 \cdot 3 ^{n} + 15}{5}$
$S_{n} = \frac{12 \cdot 6 ^{n} - 15 \cdot 3 ^{n} + 3}{5}$
Factor out 3:
$S_{n} = \frac{3}{5} (4 \cdot 6^{n} - 5 \cdot 3^{n} + 1)$

Comparing with the Given Expression

Given: $S_{n} = \frac{1}{5} (n^{2} - 12 n + 39) (4 \cdot 6^{n} - 5 \cdot 3^{n} + 1)$
Calculated: $S_{n} = \frac{3}{5} (4 \cdot 6^{n} - 5 \cdot 3^{n} + 1)$
Comparing the two:
$\frac{1}{5} (n^{2} - 12 n + 39) = \frac{3}{5}$

Solving for $n$

$n^{2} - 12 n + 39 = 3$
Subtract 3 from both sides:
$n^{2} - 12 n + 36 = 0$
Recognize the perfect square:
$(n - 6)^{2} = 0$
Solving for $n$ :
$n = 6$
Key Takeaway: Dividing by a power of a constant simplifies linear recurrences.

00:00 / 00:00

The Sigma Insight: Sum of Special Series

Analyzing the Setup

We are given the initial term

T_{1} = 6

and the recurrence relation

T_{r} = 3 T_{r - 1} + 6^{r}

. To simplify this structure, we perform a normalization by dividing both sides of the equation by

3^{r}

This yields the following transformation:

\frac{T _{r}}{3 ^{r}} = \frac{3 T _{r - 1}}{3 ^{r}} + \frac{6 ^{r}}{3 ^{r}}

Simplifying the terms, we obtain:

\frac{T _{r}}{3 ^{r}} = \frac{T _{r - 1}}{3 ^{r - 1}} + 2^{r}

The Telescoping Transformation

Let us define a new sequence

b_{r} = \frac{T _{r}}{3 ^{r}}

. The recurrence relation now takes the form of a simple difference equation:

b_{r} - b_{r - 1} = 2^{r}

To find the general term

b_{n}

, we sum these differences from

r = 2

n

b_{n} = b_{1} + r = 2 \sum n 2^{r}

Given

T_{1} = 6

, we calculate

b_{1} = \frac{6}{3} = 2

. The summation

\sum_{r = 2}^{n} 2^{r}

represents a geometric progression with first term

4

, common ratio

2

, and

n - 1

terms.

The sum is calculated as:

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

Substituting this back into our expression for

b_{n}

b_{n} = 2 + (2^{n + 1} - 4) = 2^{n + 1} - 2

Finding the General Term and the Sum

With

b_{n}

determined, we revert to the original sequence

T_{n}

T_{n} = 3^{n} \cdot b_{n} = 3^{n} (2^{n + 1} - 2) = 2 \cdot 6^{n} - 2 \cdot 3^{n}

Next, we calculate the sum

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

S_{n} = 2 r = 1 \sum n 6^{r} - 2 r = 1 \sum n 3^{r}

Evaluating these geometric series, we get:

S_{n} = 2 \cdot \frac{6 ( 6 ^{n} - 1 )}{6 - 1} - 2 \cdot \frac{3 ( 3 ^{n} - 1 )}{3 - 1}

S_{n} = \frac{12}{5} (6^{n} - 1) - 3 (3^{n} - 1)

Simplifying the expression, we arrive at:

S_{n} = \frac{3}{5} (4 \cdot 6^{n} - 5 \cdot 3^{n} + 1)

The Final Comparison

The problem provides the form

S_{n} = \frac{1}{5} (n^{2} - 12 n + 39) (4 \cdot 6^{n} - 5 \cdot 3^{n} + 1)

. By comparing our derived result with the given expression, we identify the equality:

\frac{1}{5} (n^{2} - 12 n + 39) = \frac{3}{5}

This implies:

n^{2} - 12 n + 39 = 3 \Rightarrow n^{2} - 12 n + 36 = 0

Factoring the quadratic equation, we find:

(n - 6)^{2} = 0

Thus, the final solution is $n = 6$ .

Common Traps & FAQs

Why did we divide by

3^{r}

? It feels like a random trick.

It is not random! In linear recurrence relations of the form

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

, dividing by

a^{r}

is a standard method to eliminate the coefficient of

T_{r - 1}

. It forces the recurrence into a telescoping form, which is much easier to sum.

Can we solve this using mathematical induction?

While you could verify the answer for small

n

using induction, it is not the most efficient way to derive the expression. The algebraic method of normalization and summation is more robust and provides deeper insight into the structure of the series.

Why does the summation for

b_{n}

start from

r = 2

The recurrence

T_{r} = 3 T_{r - 1} + 6^{r}

is defined for

r = 2, 3, \dots, n

. Therefore, the difference

b_{r} - b_{r - 1} = 2^{r}

is also valid for

r \geq 2

. We start from

b_{1}

and add the differences from

r = 2

up to

n

Enter Numerical Value:

Visualized Solution

The Recurrence Relation

Defining a New Sequence $b_{r}$

Calculating $b_{n}$ using Summation

Evaluating the G.P. Sum

Finding the General Term $T_{n}$

Setting up the Sum $S_{n}$

Summing the Individual G.P.s

Combining and Simplifying $S_{n}$

Comparing with the Given Expression

Solving for $n$

The Sigma Insight: Sum of Special Series

Analyzing the Setup

The Telescoping Transformation

Finding the General Term and the Sum

The Final Comparison

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

If $\frac{1 ^{3} + 2 ^{3} + 3 ^{3} + \dots upto n terms}{1 \cdot 3 + 2 \cdot 5 + 3 \cdot 7 + \dots upto n terms} = \frac{9}{5}$ , then the value of $n$ 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 the series $\sum_{n = 1}^{\infty} \frac{n ^{2} + 6 n + 10}{( 2 n + 1 )!}$ is equal to :

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

Let $S_{n} = 1 \cdot (n - 1) + 2 \cdot (n - 2) + 3 \cdot (n - 3) + \dots + (n - 1) \cdot 1, n \geq 4$ . The sum $\sum_{n = 4}^{\infty} (\frac{2 S _{n}}{n !} - \frac{1}{( n - 2 )!})$ is equal to :

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 :

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:

Let $⟨ a_{n} ⟩$ be a sequence such that $a_{1} + a_{2} + \dots + a_{n} = \frac{n ^{2} + 3 n}{( n + 1 ) ( n + 2 )}$ . If $28 \sum_{k = 1}^{10} \frac{1}{a _{k}} = p_{1} p_{2} p_{3} \dots p_{m}$ , where $p_{1}, p_{2}, \dots, p_{m}$ are the first $m$ prime numbers, then $m$ is equal to

The sum $\sum_{n = 1}^{21} \frac{3}{( 4 n - 1 ) ( 4 n + 3 )}$ is equal to

Enter Numerical Value:

Visualized Solution

The Recurrence Relation

Defining a New Sequence .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } br​

Calculating .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } bn​ using Summation

Evaluating the G.P. Sum

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

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

Summing the Individual G.P.s

Combining and Simplifying .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Sn​

Comparing with the Given Expression

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

The Sigma Insight: Sum of Special Series

Analyzing the Setup

The Telescoping Transformation

Finding the General Term and the Sum

The Final Comparison

Similar Questions

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Sum of the first n terms of the series 21​+43​+87​+1615​+… is equal to

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } If 1⋅3+2⋅5+3⋅7+…upto n terms13+23+33+…upto n terms​=59​, then the value of n is

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } For k∈N, if the sum of the series 1+k4​+k28​+k313​+k419​+… is 10, then the value of k is

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } The sum of the series ∑n=1∞​(2n+1)!n2+6n+10​ is equal to :

.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

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Let Sn​=1⋅(n−1)+2⋅(n−2)+3⋅(n−3)+⋯+(n−1)⋅1,n≥4. The sum ∑n=4∞​(n!2Sn​​−(n−2)!1​) is equal to :

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } If the sum of the first 15 terms of the series (3/4)3+(1/2)3+(2/3)3+33+(3/4)3+… is equal to 225k, then k is equal to :

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } If the sum of the first ten terms of the series (153​)2+(252​)2+(351​)2+42+(454​)2+… is 516​m, then m is equal to:

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } The sum ∑n=121​(4n−1)(4n+3)3​ is equal to

Defining a New Sequence $b_{r}$

Calculating $b_{n}$ using Summation

Finding the General Term $T_{n}$

Setting up the Sum $S_{n}$

Combining and Simplifying $S_{n}$

Solving for $n$

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

If $\frac{1 ^{3} + 2 ^{3} + 3 ^{3} + \dots upto n terms}{1 \cdot 3 + 2 \cdot 5 + 3 \cdot 7 + \dots upto n terms} = \frac{9}{5}$ , then the value of $n$ 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 the series $\sum_{n = 1}^{\infty} \frac{n ^{2} + 6 n + 10}{( 2 n + 1 )!}$ is equal to :

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

Let $S_{n} = 1 \cdot (n - 1) + 2 \cdot (n - 2) + 3 \cdot (n - 3) + \dots + (n - 1) \cdot 1, n \geq 4$ . The sum $\sum_{n = 4}^{\infty} (\frac{2 S _{n}}{n !} - \frac{1}{( n - 2 )!})$ is equal to :

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 :

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 $\sum_{n = 1}^{21} \frac{3}{( 4 n - 1 ) ( 4 n + 3 )}$ is equal to