{ "@context": "https://schema.org/", "@type": "Quiz", "about": { "@type": "Thing", "name": "Summation" }, "educationalAlignment": [ { "@type": "AlignmentObject", "alignmentType": "educationalSubject", "targetName": "Mathematics" } ], "hasPart": [ { "@context": "https://schema.org/", "@type": "Question", "eduQuestionType": "Flashcard", "text": "Let {a_n}_n=1 be a sequence such that a_1=1,a_2=1 and a_(n+2)=2a_(n+1) + a_n for all n>=1. Then the value of 47S{a_n}/{8^n} is equal to", "acceptedAnswer": { "@type": "Answer", "text": "Given, a_{n+2} = 2a_{n+1} + a_n => Dividing both sides by 8^n => Therefore, {a_{n+2}}/{8^n} = {2a_{n + 1}}/{8^n} + {a_n}/{8^n} => 64 * {a_{n+2}}/{8^{n+2}} = 16 * {a_{n + 1}}/{8^n} + {a_n}/{8^n} Applying Summation to both sides we get, 64*S_{n=1}^{infinity} {a_{n+2}}/{8^{n+2}}=16*S_{n=1}^{infinity}{a_{n+1}}/{8^{n+1}} + S_{n=1}^{infinity}{a_n}/{8^n} Let s = S_{n=1}^{infinity}{a_n}/{8^n}then S_{n=1}^{infinity}{a_{n+2}}/{8^{n+2}} = s - {a_1}/{8} - {a_2}/{64} S_{n=1}^{infinity}{a_{n + 1}}/{8^{n+1}} = s - {a_1}{8} Therefore from above equation 64(s - {a_1}/{8} - {a_2}/{64}) = 16(s - {a_1}/{8}) + s => 64s - 8 - 1 = 17s - 2 => 47s = 7 Therefore 47S_{n=1}^{infinity}{a_n}/{8^n} = 7" } } ] } { "@context": "http://schema.org", "@type": "QAPage", "name": "Let {a_n}_n=1 be a sequence such that a_1=1,a_2=1 and a_(n+2)=2a_(n+1) + a_n for all n>=1. Then the value of 47S{a_n}/{8^n} is equal to", "description": "Given, a_{n+2} = 2a_{n+1} + a_n => Dividing both sides by 8^n => Therefore, {a_{n+2}}/{8^n} = {2a_{n + 1}}/{8^n} + {a_n}/{8^n} => 64 * {a_{n+2}}/{8^{n+2}} = 16 * {a_{n + 1}}/{8^n} + {a_n}/{8^n}...", "mainEntity": { "@type": "Question", "@id": "let-a_n_n1-be-a-sequence-such-that-a_11a_21-and-a_n22a_n1--a_n-for-all-n1-then-the-value-of-47sa_n-8n-is-equal-to/", "name": "Let {a_n}_n=1 be a sequence such that a_1=1,a_2=1 and a_(n+2)=2a_(n+1) + a_n for all n>=1. Then the value of 47S{a_n}/{8^n} is equal to", "text": "Let {a_n}_n=1 be a sequence such that a_1=1,a_2=1 and a_(n+2)=2a_(n+1) + a_n for all n>=1. Then the value of 47S{a_n}/{8^n} is equal to", "dateCreated": "2023-01-19T14:20Z", "answerCount": "1", "author": { "@type": "Person", "name": "ReadAxis", "url": "https://www.readaxis.com/" }, "acceptedAnswer": { "@type": "Answer", "upvoteCount": "1", "text": "Given, a_{n+2} = 2a_{n+1} + a_n => Dividing both sides by 8^n => Therefore, {a_{n+2}}/{8^n} = {2a_{n + 1}}/{8^n} + {a_n}/{8^n} => 64 * {a_{n+2}}/{8^{n+2}} = 16 * {a_{n + 1}}/{8^n} + {a_n}/{8^n} Applying Summation to both sides we get, 64*S_{n=1}^{infinity} {a_{n+2}}/{8^{n+2}}=16*S_{n=1}^{infinity}{a_{n+1}}/{8^{n+1}} + S_{n=1}^{infinity}{a_n}/{8^n} Let s = S_{n=1}^{infinity}{a_n}/{8^n}then S_{n=1}^{infinity}{a_{n+2}}/{8^{n+2}} = s - {a_1}/{8} - {a_2}/{64} S_{n=1}^{infinity}{a_{n + 1}}/{8^{n+1}} = s - {a_1}{8} Therefore from above equation 64(s - {a_1}/{8} - {a_2}/{64}) = 16(s - {a_1}/{8}) + s => 64s - 8 - 1 = 17s - 2 => 47s = 7 Therefore 47S_{n=1}^{infinity}{a_n}/{8^n} = 7.", "url": "let-a_n_n1-be-a-sequence-such-that-a_11a_21-and-a_n22a_n1--a_n-for-all-n1-then-the-value-of-47sa_n-8n-is-equal-to/", "dateCreated": "2023-01-19T14:20Z", "author": { "@type": "Person", "name": "Readaxis Admin" } }, "suggestedAnswer": [] } }

Question

Let \(\{a_n\}_{n=1}\) be a sequence such that \(a_1 = 1 , a_2 = 1\) and \(a_{n + 2} = 2a_{n + 1} + a_n\) for all \(n \geq 1\). Then the value of \[47 \sum_{n=1}^{\infty} \frac{a_n}{8^n}\] is equal to

Solution:

Given, \(a_{n+2} = 2a_{n+1} + a_n\)
Dividing both sides by \(8^n\)

Therefore, \(\frac{a_{n+2}}{8^n} = \frac{2a_{n + 1}}{8^n} + \frac{a_n}{8^n}\)
\( \Rightarrow 64 \times \frac{a_{n+2}}{8^{n+2}} = 16 \times \frac{a_{n + 1}}{8^n} + \frac{a_n}{8^n}\)

Applying Summation to both sides

we get, \( 64 \sum_{n=1}^{\infty} \frac{a_{n+2}}{8^{n+2}} = 16 \sum_{n=1}^{\infty}\frac{a_{n + 1}}{8^{n+1}} + \sum_{n=1}^{\infty}\frac{a_n}{8^n}\)

Let \( S = \sum_{n=1}^{\infty}\frac{a_n}{8^n}\)
then \(\sum_{n=1}^{\infty} \frac{a_{n+2}}{8^{n+2}} = S – \frac{a_1}{8} – \frac{a_2}{64}\)
\(\sum_{n=1}^{\infty}\frac{a_{n + 1}}{8^{n+1}} = S – \frac{a_1}{8} \)

Therefore from above equation
\(64(S – \frac{a_1}{8} – \frac{a_2}{64}) = 16(S – \frac{a_1}{8} ) + S \)
\( \Rightarrow 64S – 8 – 1 = 17S – 2\)
\( \Rightarrow 47S = 7\)

Therefore \[47\sum_{n=1}^{\infty}\frac{a_n}{8^n} = 7\]