Start free

Mathematics · Sequence and Series

JEE Advanced 2019 — Paper 1 — Question 33

Let α\alpha and β\beta be the roots of x2−x−1=0x^{2}-x-1=0, with α>β\alpha>\beta. For all positive integers nn, define

an=an−βnα−β,n≥1,b1=1a_{n}=\frac{a^{n}-\beta^{n}}{\alpha-\beta}, \quad n \geq 1, b_{1}=1 and bn=an−1+an+1,n≥2b_{n}=a_{n-1}+a_{n+1}, n \geq 2. Then which of the following

options is/are correct?

  1. Option A:

    ∑n=1∞bn10n=889\sum_{\mathrm{n}=1}^{\infty} \frac{\mathrm{b}_{\mathrm{n}}}{10^{\mathrm{n}}}=\frac{8}{89}

  2. Option B:

    bn=αn+βnb_{n}=\alpha^{n}+\beta^{n} for all n≥1n \geq 1

    Correct
  3. Option C:

    a1+a2+a3+…..+an=an+2−1a_{1}+a_{2}+a_{3}+\ldots . .+a_{n}=a_{n+2}-1 for all n≥1n \geq 1

    Correct
  4. Option D:

    ∑n=1∞an10n=1089\sum_{\mathrm{n}=1}^{\infty} \frac{\mathrm{a}_{\mathrm{n}}}{10^{\mathrm{n}}}=\frac{10}{89}

    Correct

Answer: B, C, D

Step-by-step solution

Clearly we have α+β=1&αβ=−1\alpha+\beta=1 \& \alpha \beta=-1

α,β=1±52 As bn=an−1+an+1=αn−1−βn−1α−β+αn+1−βn+1α−β\begin{gathered} \alpha, \beta=\frac{1 \pm \sqrt{5}}{2} \text { As } b_{n}=a_{n-1}+a_{n+1} =\frac{\alpha^{n-1}-\beta^{n-1}}{\alpha-\beta}+\frac{\alpha^{n+1}-\beta^{n+1}}{\alpha-\beta} \end{gathered} =αn−1(1+α2)−βn−1(1+β2)α−β=αn−1(α+2)−βn−1(β+2)α−β=αn−1(5+52)−βn−1(5−52)α−β=5(αn+βn)α−β=αn+βn: As α−β=5 (D) ∑n=1∞an10n=∑n=1∞αn−βn(α−β)10n=α101−α10−β101−β10(α−β)=α(10−β)−β(10−α)(10−α)(10−β)(α−β)=1089 (A) ∑n=1∞bn10n=∑n=1∞αn+βn10n=α101−α10+β101−β10=α(10−β)+β(10−α)(10−α)(10−β)=10(α+β)−2αβ100−(α+β)10+αβ=1289 (C) a1+a2+a3+…+an=∑r=1nar=∑r=1nαr−βrα−β=α(1−αn)1−α−β(1−βn)1−βα−β=(α−αβ)(1−αn)−β(1−α)(1−βn)(α−β)(1−α)(1−β)=(α−β)−αn(1+α)+βn(1+β)−(α−β)=(α−β)−αn+2+βn+2−(α−β); As 1+x=x2=−1+αn+2−βn+2α−β=−1+an+2\begin{aligned} & =\frac{\alpha^{\mathrm{n}-1}\left(1+\alpha^{2}\right)-\beta^{\mathrm{n}-1}\left(1+\beta^{2}\right)}{\alpha-\beta} \\& =\frac{\alpha^{\mathrm{n}-1}(\alpha+2)-\beta^{\mathrm{n}-1}(\beta+2)}{\alpha-\beta} \\& =\frac{\alpha^{\mathrm{n}-1}\left(\frac{5+\sqrt{5}}{2}\right)-\beta^{\mathrm{n}-1}\left(\frac{5-\sqrt{5}}{2}\right)}{\alpha-\beta} \\& =\frac{\sqrt{5}\left(\alpha^{\mathrm{n}}+\beta^{\mathrm{n}}\right)}{\alpha-\beta}=\alpha^{\mathrm{n}}+\beta^{\mathrm{n}}: \text { As } \alpha-\beta=\sqrt{5} \\& \text { (D) } \sum_{\mathrm{n}=1}^{\infty} \frac{\mathrm{a}_{\mathrm{n}}}{10^{\mathrm{n}}}=\sum_{\mathrm{n}=1}^{\infty} \frac{\alpha^{\mathrm{n}}-\beta^{\mathrm{n}}}{(\alpha-\beta) 10^{\mathrm{n}}} \\& =\frac{\frac{\frac{\alpha}{10}}{1-\frac{\alpha}{10}}-\frac{\frac{\beta}{10}}{1-\frac{\beta}{10}}}{(\alpha-\beta)} \\& =\frac{\alpha(10-\beta)-\beta(10-\alpha)}{(10-\alpha)(10-\beta)(\alpha-\beta)}=\frac{10}{89} \\& \text { (A) } \sum_{\mathrm{n}=1}^{\infty} \frac{\mathrm{b}_{\mathrm{n}}}{10^{\mathrm{n}}}=\sum_{\mathrm{n}=1}^{\infty} \frac{\alpha^{\mathrm{n}}+\beta^{\mathrm{n}}}{10^{\mathrm{n}}}=\frac{\frac{\alpha}{10}}{1-\frac{\alpha}{10}}+\frac{\frac{\beta}{10}}{1-\frac{\beta}{10}} \\& =\frac{\alpha(10-\beta)+\beta(10-\alpha)}{(10-\alpha)(10-\beta)} \\& =\frac{10(\alpha+\beta)-2 \alpha \beta}{100-(\alpha+\beta) 10+\alpha \beta}=\frac{12}{89} \\& \text { (C) } \mathrm{a}_{1}+\mathrm{a}_{2}+\mathrm{a}_{3}+\ldots+\mathrm{a}_{\mathrm{n}}=\sum_{\mathrm{r}=1}^{\mathrm{n}} \mathrm{a}_{\mathrm{r}}=\sum_{\mathrm{r}=1}^{\mathrm{n}} \frac{\alpha^{\mathrm{r}}-\beta^{\mathrm{r}}}{\alpha-\beta} \\& =\frac{\frac{\alpha\left(1-\alpha^{\mathrm{n}}\right)}{1-\alpha}-\frac{\beta\left(1-\beta^{\mathrm{n}}\right)}{1-\beta}}{\alpha-\beta} \\& =\frac{(\alpha-\alpha \beta)\left(1-\alpha^{\mathrm{n}}\right)-\beta(1-\alpha)\left(1-\beta^{\mathrm{n}}\right)}{(\alpha-\beta)(1-\alpha)(1-\beta)} \\& =\frac{(\alpha-\beta)-\alpha^{\mathrm{n}}(1+\alpha)+\beta^{\mathrm{n}}(1+\beta)}{-(\alpha-\beta)} \\& =\frac{(\alpha-\beta)-\alpha^{n+2}+\beta^{n+2}}{-(\alpha-\beta)} ; \text { As } 1+x=x^{2} \\& =-1+\frac{\alpha^{n+2}-\beta^{n+2}}{\alpha-\beta}=-1+a_{n+2} \end{aligned}

Answer key and solution verified before publishing.

Practise Sequence and Series

Start with this question, then two more from the same chapter — with a tutor that explains every step. Free.

Solve a similar one free
Exam
JEE Advanced 2019
Paper
Paper 1
Subject
Mathematics
Chapter
Sequence and Series
Topic
Geometric Progression

More Sequence and Series questions from this paper

Let α and β be the roots of x 2 -x-1=0 , with α β . For all positive… | JEE Advanced 2019 PYQ with Solution · DhiX AI