JEE Main 2021 (20 July Shift 2)

LEVELJEE Main

Animated Solution for Mathematics - Sequence and Series: Let ${a_{n}}_{n = 1}^{\infty}$ be a sequence such that $a_{1} = 1, a_{2} = 1$ and $a_{n + 2} = 2 a_{n + 1} + a_{n}$ for all $n \geq 1$ . Then the value of $47 \sum_{n = 1}^{\infty} \frac{a _{n}}{2 ^{3 n}}$ is equal to

Q: Wait, why do we subtract terms like $\frac{a_1}{8}$? I thought the sum was just $P$.

The sum $P$ is defined from $n=1$. If you shift the index to $n+1$, the new sum starts from $n=2$. To keep it equal to $P$, you must subtract the term that was 'missing' or 'extra' compared to the original series. It is all about keeping the bounds of the summation consistent.

Q: Is it possible that this series diverges?

Great question! For this specific recurrence, $a_n$ grows roughly like $(\sqrt{2}+1)^n$. Since $\sqrt{2}+1 \approx 2.414$, which is much less than $8$, the terms $\frac{a_n}{8^n}$ decay rapidly, ensuring the series converges beautifully.

Enter Numerical Value:

Visualized Solution

Identify the Sequence

Given sequence: $a_{1} = 1, a_{2} = 1$
Recurrence relation: $a_{n + 2} = 2 a_{n + 1} + a_{n}$ for $n \geq 1$

Analyze the Target Sum

Target: $47 \sum_{n = 1}^{\infty} \frac{a _{n}}{2 ^{3 n}}$
Simplify denominator: $2^{3 n} = (2^{3})^{n} = 8^{n}$
Let $P = \sum_{n = 1}^{\infty} \frac{a _{n}}{8 ^{n}}$

Transform the Recurrence

Divide recurrence by $8^{n}$ :
$\frac{a _{n + 2}}{8 ^{n}} = \frac{2 a _{n + 1}}{8 ^{n}} + \frac{a _{n}}{8 ^{n}}$

Adjusting the Powers

Adjust denominators to match indices:
$\frac{64 a _{n + 2}}{8 ^{n + 2}} = \frac{16 a _{n + 1}}{8 ^{n + 1}} + \frac{a _{n}}{8 ^{n}}$

Apply Summation

Summing from $n = 1$ to $\infty$ :
$64 \sum_{n = 1}^{\infty} \frac{a _{n + 2}}{8 ^{n + 2}} = 16 \sum_{n = 1}^{\infty} \frac{a _{n + 1}}{8 ^{n + 1}} + \sum_{n = 1}^{\infty} \frac{a _{n}}{8 ^{n}}$

Relating to $P$

Express in terms of $P$ :
$\sum_{n = 1}^{\infty} \frac{a _{n}}{8 ^{n}} = P$
$\sum_{n = 1}^{\infty} \frac{a _{n + 1}}{8 ^{n + 1}} = P - \frac{a _{1}}{8}$
$\sum_{n = 1}^{\infty} \frac{a _{n + 2}}{8 ^{n + 2}} = P - \frac{a _{1}}{8} - \frac{a _{2}}{8 ^{2}}$

Substitute Known Values

Substitute $a_{1} = 1, a_{2} = 1$ :
$64 (P - \frac{1}{8} - \frac{1}{64}) = 16 (P - \frac{1}{8}) + P$

Expand the Equation

Expanding both sides:
$64 P - 8 - 1 = 16 P - 2 + P$

Simplify Both Sides

Combine like terms:
$64 P - 9 = 17 P - 2$

Solve for $P$

Rearranging terms:
$64 P - 17 P = 9 - 2$
$47 P = 7$

00:00 / 00:00

The Sigma Insight: Sum of Special Series

Solution Diagram

Analyzing the Setup

We are given a sequence defined by

a_{1} = 1

a_{2} = 1

, and the recurrence relation

a_{n + 2} = 2 a_{n + 1} + a_{n}

. Our objective is to evaluate the sum:

47 n = 1 \sum \infty \frac{a _{n}}{2 ^{3 n}}

Recognizing that

2^{3 n} = (2^{3})^{n} = 8^{n}

, we can rewrite the expression as

47 \sum_{n = 1}^{\infty} \frac{a _{n}}{8 ^{n}}

. Let us define our target sum as

P

P = n = 1 \sum \infty \frac{a _{n}}{8 ^{n}}

The Art of Manipulation

We begin with the recurrence relation

a_{n + 2} = 2 a_{n + 1} + a_{n}

. To align this with our sum

P

, we divide the entire equation by

8^{n}

\frac{a _{n + 2}}{8 ^{n}} = \frac{2 a _{n + 1}}{8 ^{n}} + \frac{a _{n}}{8 ^{n}}

To match the indices with the powers of

8

, we multiply the terms to obtain powers of

8^{n + 2}

8^{n + 1}

, and

8^{n}

respectively:

\frac{64 a _{n + 2}}{8 ^{n + 2}} = \frac{16 a _{n + 1}}{8 ^{n + 1}} + \frac{a _{n}}{8 ^{n}}

The Shifting Technique

Now, we apply the summation

\sum_{n = 1}^{\infty}

to both sides of the equation. We express the shifted sums in terms of

P

n = 1 \sum \infty \frac{a _{n + 1}}{8 ^{n + 1}} = P - \frac{a _{1}}{8}

n = 1 \sum \infty \frac{a _{n + 2}}{8 ^{n + 2}} = P - \frac{a _{1}}{8} - \frac{a _{2}}{8 ^{2}}

Substituting these into our summation equation, we obtain:

64 (P - \frac{a _{1}}{8} - \frac{a _{2}}{8 ^{2}}) = 16 (P - \frac{a _{1}}{8}) + P

The Algebraic Climax

Given

a_{1} = 1

and

a_{2} = 1

, we substitute these values into the equation:

64 (P - \frac{1}{8} - \frac{1}{64}) = 16 (P - \frac{1}{8}) + P

Expanding the terms yields:

64 P - 8 - 1 = 16 P - 2 + P

64 P - 9 = 17 P - 2

Solving for

P

, we find

47 P = 7

. Since our original goal was to evaluate

47 P

, the final result is:

Common Traps & FAQs

Why do we divide by

8^{n}

and not something else?

We want the denominator to match the index of the sequence term. Since we have

a_{n + 2}

, we want

8^{n + 2}

in the denominator. Dividing by

8^{n}

is the first step to creating that structure, allowing us to relate the shifted terms back to our original sum

P

Wait, why do we subtract terms like

\frac{a _{1}}{8}

? I thought the sum was just

P

The sum

P

is defined from

n = 1

. If you shift the index to

n + 1

, the new sum starts from

n = 2

. To keep it equal to

P

, you must subtract the term that was 'missing' or 'extra' compared to the original series. It is all about keeping the bounds of the summation consistent.

Is it possible that this series diverges?

Great question! For this specific recurrence,

a_{n}

grows roughly like

(2 + 1)^{n}

. Since

2 + 1 \approx 2.414

, which is much less than

8

, the terms

\frac{a _{n}}{8 ^{n}}

decay rapidly, ensuring the series converges beautifully.

Animated Solution for Mathematics - Sequence and Series: Let ${a_{n}}_{n = 1}^{\infty}$ be a sequence such that $a_{1} = 1, a_{2} = 1$ and $a_{n + 2} = 2 a_{n + 1} + a_{n}$ for all $n \geq 1$ . Then the value of $47 \sum_{n = 1}^{\infty} \frac{a _{n}}{2 ^{3 n}}$ is equal to

Enter Numerical Value:

Visualized Solution

Identify the Sequence

Analyze the Target Sum

Transform the Recurrence

Adjusting the Powers

Apply Summation

Relating to $P$

Substitute Known Values

Expand the Equation

Simplify Both Sides

Solve for $P$

The Sigma Insight: Sum of Special Series

Analyzing the Setup

The Art of Manipulation

The Shifting Technique

The Algebraic Climax

Similar Questions

Let ${a_{n}}_{n = 0}^{\infty}$ be a sequence such that $a_{0} = a_{1} = 0$ and $a_{n + 2} = 2 a_{n + 1} - a_{n} + 1$ for all $n \geq 0$ . Then, $\sum_{n = 2}^{\infty} \frac{a _{n}}{7 ^{n}}$ is equal to

Let $⟨ a_{n} ⟩$ be a sequence such that $a_{0} = 0$ , $a_{1} = \frac{1}{2}$ and $2 a_{n + 2} = 5 a_{n + 1} - 3 a_{n}$ , $n = 0, 1, 2, 3, ...$ Then $\sum_{k = 1}^{100} a_{k}$ is

Let $S_{k}, k = 1, 2, \dots, 100$ , denote the sum of the infinite geometric series whose first term is $\frac{k - 1}{k !}$ and the common ratio is $\frac{1}{k}$ . Then the value of $\frac{10 0 ^{2}}{100 !} + \sum_{k = 1}^{100} ∣ (k^{2} - 3 k + 1) S_{k} ∣$ is \dots.

Let ${a_{n}}_{n = 0}^{\infty}$ be a sequence such that $a_{0} = a_{1} = 0$ and $a_{n + 2} = 3 a_{n + 1} - 2 a_{n} + 1, \forall n \geq 0$ . Then $a_{25} a_{23} - 2 a_{25} a_{22} - 2 a_{23} a_{24} + 4 a_{22} a_{24}$ is equal to:

The sum, $\sum_{n = 1}^{7} \frac{n ( n + 1 ) ( 2 n + 1 )}{4}$ is equal to

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The sum, $\sum_{n = 1}^{7} \frac{n ( n + 1 ) ( 2 n + 1 )}{4}$ is equal to _________ .

If $a_{n} = \frac{- 2}{4 n ^{2} - 16 n + 15}$ , then $a_{1} + a_{2} + \dots + a_{25}$ 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

Let $a_{1} = b_{1} = 1$ and $a_{n} = a_{n - 1} + (n - 1), b_{n} = b_{n - 1} + a_{n - 1}, \forall n \geq 2$ . If $S = \sum_{n = 1}^{10} \frac{b _{n}}{2 ^{n}}$ and $T = \sum_{n = 1}^{8} \frac{n}{2 ^{n - 1}}$ , then $2^{7} (2 S - T)$ is equal to

Enter Numerical Value:

Visualized Solution

Identify the Sequence

Analyze the Target Sum

Transform the Recurrence

Adjusting the Powers

Apply Summation

Relating to .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } P

Substitute Known Values

Expand the Equation

Simplify Both Sides

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

The Sigma Insight: Sum of Special Series

Analyzing the Setup

The Art of Manipulation

The Shifting Technique

The Algebraic Climax

Similar Questions

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.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } If an​=4n2−16n+15−2​, then a1​+a2​+⋯+a25​ is equal to :

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Let a1​=b1​=1 and an​=an−1​+(n−1),bn​=bn−1​+an−1​,∀n≥2. If S=∑n=110​2nbn​​ and T=∑n=18​2n−1n​, then 27(2S−T) is equal to

Relating to $P$

Solve for $P$

Let ${a_{n}}_{n = 0}^{\infty}$ be a sequence such that $a_{0} = a_{1} = 0$ and $a_{n + 2} = 2 a_{n + 1} - a_{n} + 1$ for all $n \geq 0$ . Then, $\sum_{n = 2}^{\infty} \frac{a _{n}}{7 ^{n}}$ is equal to

Let $⟨ a_{n} ⟩$ be a sequence such that $a_{0} = 0$ , $a_{1} = \frac{1}{2}$ and $2 a_{n + 2} = 5 a_{n + 1} - 3 a_{n}$ , $n = 0, 1, 2, 3, ...$ Then $\sum_{k = 1}^{100} a_{k}$ is

The sum, $\sum_{n = 1}^{7} \frac{n ( n + 1 ) ( 2 n + 1 )}{4}$ is equal to

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, $\sum_{n = 1}^{7} \frac{n ( n + 1 ) ( 2 n + 1 )}{4}$ is equal to _________ .

If $a_{n} = \frac{- 2}{4 n ^{2} - 16 n + 15}$ , then $a_{1} + a_{2} + \dots + a_{25}$ is equal to :

Let $a_{1} = b_{1} = 1$ and $a_{n} = a_{n - 1} + (n - 1), b_{n} = b_{n - 1} + a_{n - 1}, \forall n \geq 2$ . If $S = \sum_{n = 1}^{10} \frac{b _{n}}{2 ^{n}}$ and $T = \sum_{n = 1}^{8} \frac{n}{2 ^{n - 1}}$ , then $2^{7} (2 S - T)$ is equal to