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The following problem is taken from JEE exam practice set.

Let $\{a_n\}_{n=1}^{\infty}$ be a sequence such that $a_1=1,a_2=1$ and $a_{n+2}=2a_{n+1}+a_{n}$ for all $n\ge1$. Find the value of $47\sum_{n=1}^{\infty}\dfrac{a_n}{2^{3n}}$

Attempt$1$: The given sequence is $1,1,3,7,17,41,98...$

Let $S=\dfrac{1}{2^3}+\dfrac{1}{2^6}+\dfrac{3}{2^9}+\dfrac{7}{2^{12}}+\dfrac{41}{2^{15}}+\dfrac{98}{2^{18}}+...$

$\dfrac{S}{2^3}=\dfrac{1}{2^6}+\dfrac{1}{2^9}+\dfrac{3}{2^{12}}+\dfrac{7}{2^{15}}+\dfrac{41}{2^{18}}+\dfrac{98}{2^{21}}+...$

$\implies(1-\dfrac1{2^3})S=\dfrac{1}{2^3}+\dfrac{2}{2^9}+\dfrac{4}{2^{12}}+\dfrac{34}{2^{15}}+\dfrac{57}{2^{18}}+...$

Not able to finish.

Attempt$2$: $a_3=2+1, a_4=2^2+2+1, a_5=2^3+2^2+2+2+1=2^3+2^2+2^2+1=2^4+1$

$a_6=2^5+2+2^2+2+1=2^5+2^3+1, a_7=2^6+2^4+2+2^4+1=2^6+2^5+2+1$

Not able to finish this one either.

Attempt$3$: $\dfrac{a_n}{2^{3n}}=\dfrac{a_{n+2}-2a_{n+1}}{2^{3n}}=\dfrac{a_{n+2}}{2^{3n}}-\dfrac{a_{n+1}}{2^{3n-1}}$

Don't know what I can do with this one.

Attempt$4$:$\dfrac{a_n}{2^{3n}}=\dfrac{2{a_{n-1}}+a_{n-2}}{2^{3n}}=\dfrac{2^2a_{n-2}+2a_{n-3}+a_{n-2}}{2^{3n}}=\dfrac{(2^2+1)a_{n-2}+a_{n-3}}{2^{3n}}=\dfrac{(2^3+2+1)a_{n-3}+(2^2+1)a_{n-4}}{2^{3n}}$

Not feeling confident with this one either.

aarbee
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4 Answers4

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We use the recursion relation itself to form appropriate series.

$$S=\frac{1}{8} + \frac{1}{8^2} + \frac{3}{8^3} + \frac{7}{8^4} + \frac{17}{8^5} + \ldots$$

The relation given is $a_{n}=a_{n+2}-2a_{n+1}$. Let us write $$\frac{2}{8}S= 0 + \frac{2\cdot 1}{8^2} + \frac{2\cdot 1}{8^3} + \frac{2 \cdot 3}{8^4} + \frac{2 \cdot 7}{8^5} + \ldots$$

On subtracting we get, $$\frac{3}{4}S = \frac{1}{8} - \frac{1}{8^2} + \frac{1}{8^3} + \frac{1}{8^4} + \frac{3}{8^5} + \ldots$$ which is $$\frac{3}{4}S = \frac{1}{8} - \frac{1}{8^2} + \frac{1}{8^2}S$$ resulting in $$S=\frac{7}{47}$$

MyMolecules
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  • Hi. How did you write the $\dfrac28S$ step? Can you elaborate? Thanks. – aarbee Jan 30 '22 at 18:16
  • @aarbee Instead of simply multiplying by common ratio, I multiplied S with twice of common ratio because of $2a_{n+1}$ term in recursion relation. This allowed to create another S term on RHS and obtain a linear equation to solve. – MyMolecules Jan 31 '22 at 04:07
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You can show that for all $n \geq 1$ $$ a_n = \frac{\sqrt{2}-1}{2}\left[\left(1+\sqrt{2}\right)^n - \left(3+2\sqrt{2}\right)\left(1-\sqrt{2}\right)^n\right] $$ Do you know how to proceed ? Then $$ \sum_{n=1}^{+\infty}\frac{a_n}{2^{3n}} = \frac{\sqrt{2}-1}{2}\left(\sum_{n=1}^{+\infty}\left(\frac{1+\sqrt{2}}{8}\right)^n - \left(3+2\sqrt{2}\right)\sum_{n=1}^{+\infty}\left(\frac{1-\sqrt{2}}{8}\right)^n \right) $$

Atmos
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0

Write $F(X)=\sum_{n=1}^\infty a_n X^n$. Compute $F(X)-2XF(X)-X^2F(X)$ by observing that the coefficient of $X^n$ in this expression is $0$ for $n \ge 3$. The result is, therefore, a polynomial.

Now you have $F(X)$ represented as a rational function. Put $X=\frac{1}{8}$ to evaluate $\sum a_n (1/2)^{3n}$ and you’re done.

Alon Amit
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0

The sequence $a_{n+2}-2a_{n+1}-a_n=0$ verifies a linear recurrence relation with constant coefficients.

It has a solution $a_n=\alpha{r_1}^n+\beta{r_2}^n\ $ in our case $r_i=1\pm\sqrt{2}$, solutions of $x^2-2x-1=0$.

$\displaystyle S=\sum\limits_{n=1}^\infty\dfrac{a_n}{2^{3n}}=\alpha\sum\limits_{n=1}^\infty\left(\dfrac{r_1}{8}\right)^n+\beta\sum\limits_{n=1}^\infty\left(\dfrac{r_2}{8}\right)^n=\sum\limits_{n=1}^\infty b_n$

Where the sequence $b_n=\alpha\left(\dfrac{r_1}{8}\right)^n+\beta\left(\dfrac{r_2}{8}\right)^n$ verifies the equation:

$$b_{n+2}-\frac 28b_{n+1}-\frac 1{64}b_n=0\iff 64b_{n+2}=16b_{n+1}+b_n$$

Since $|\frac{r_1}8|<1$ and $|\frac{r_2}8|<1$, the series $S$ converge absolutely, which justifies the following manipulations:

$\begin{align}S&=b_1+\sum\limits_{n=1}^\infty b_{n+1}\\ S&=b_1+b_2+\sum\limits_{n=1}^\infty b_{n+2}\end{align}$

Therefore $64(S-b_1-b_2)=16(S-b_1)+S\iff S=\dfrac{48b_1+64b_2}{47}$

$b_1=\frac {a_1}8=\frac 18\ $ and $\ b_2=\frac {a_2}{8^2}=\frac 1{64}\implies 47S=6+1=7$

zwim
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