Series Problem | SMO, 2013 | Problem 27
Try this beautiful problem from Singapore Mathematics Problem, SMO, 2012 based on Series. You may use sequential hints to solve the problem.
Series Problem - (SMO Entrance)
Find the value of \(\lfloor(\frac {3+\sqrt {17}}{2})^{6} \rfloor \)
Key Concepts
Check the Answer
Answer : 2041
Singapore Mathematics Olympiad
Challenges and Thrills - Pre - College Mathematics
Try with Hints
If you got stuck in this hint please do through this hint :
We can assume two integers, \(\alpha\) and \(\beta\) where ,
\(\alpha = \frac {3+\sqrt {17}}{2}\) so let \(\beta = \frac {3 - \sqrt {17}}{2}\)
Then if we try to substitute the value in any formula lets try to find the value of \(\alpha . \beta\)
and \(\alpha + \beta \)
I think you can definitely find the values.........Try .......................
If you are still not getting it ..................
\(\alpha .\beta = \frac {3+\sqrt {17}}{2} . \frac {3-\sqrt {17}}{2} = -2 \)
and again \(\alpha + \beta = 3\)
Lets consider the sum \(s_{n} = \alpha ^ {n} + \beta ^{n}\).
Then try to find the value of \( 3 s_{n+1} + 2 s_{n}\) ................................
This one is the last hint ...........
\( 3 s_{n+1} + 2 s_{n} = (\ (alpha +\beta)(\alpha^{n+1} + \beta^ {n+1} -\alpha\beta(\alpha^{n} + \beta ^{n}\)
= \(\alpha ^ {n+2} - \beta ^ {n+2} = s_{n+2}\)
Note that \(|\beta| < 1\) .Then for even positive integer n, \(\lfloor a^{n}\rfloor = s_{n} + \lfloor - \beta^n \rfloor = s_{n} - 1\).
As \(s_{0} = 2\) and \(s_{1}=3\) we can find \(s_{6} = 2041\)
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Radius of Convergence of a Power series | IIT JAM 2016
Try this problem from IIT JAM 2017 exam (Problem 48) and know how to determine radius of convergence of a power series.
Radius of Convergence of a Power Series | IIT JAM 2016 | Problem 48
Find the radius of convergence of the power series
Key Concepts
Check the Answer
Answer: $\frac12$
IIT JAM 2016 , Problem 48
Real Analysis : Robert G. Bartle
Try with Hints
Given, the power series is $\sum_{n=1}^{\infty} \frac{(-4)^{n}}{n(n+1)}(x+2)^{2 n}$.
Let us put $2n=m$ to get the standard form of a power series.
We get,
$\sum_{m=2}^{\infty} \frac{(-4)^{\frac m2}}{\frac m2(\frac m2+1)}(x+2)^{ m}$.
Now let us make the transformation $z=x+2$ to get a power series about 0 :
We have,
$\sum_{m=2}^{\infty} \frac{(-4)^{\frac m2}}{\frac m2(\frac m2+1)}(z)^{ m}$
Compairing with $ \sum_{m=2}^{\infty} a_m (z)^m$
we get,
$a_m= \frac{(-4)^{\frac m2}}{\frac m2(\frac m2+1)} $
Now we have to test the convergence of the series.
Can you apply Ratio Test to check the convergence of the series.
Ratio Test : Let $\sum_{n=0}^{\infty} a_{n} x^{n}$ be a power series and let $\lim \left|\frac{a_{n+1}}{a_{n}}\right|=\mu .$ Then
if $\mu=0$ the series is everywhere convergent; if $0<\mu<\infty$ the series is absolutely convergent for all $x$ satisfyir $|x|<\frac{1}{\mu}$ and the series is divergent for all $x$ satisfying $|x|>\frac{1}{\mu}$ if $\mu=\infty,$ the series is nowhere converegnt.
$\begin{aligned}\left|\frac{a_{m+1}}{a_m}\right| &=\left| \frac{4^{\frac m2}\cdot 2\cdot4}{(m+1)(m+3)} \times \frac{m(m+2)}{4^{\frac m2} \cdot 4}\right| \\&=\left| \quad \frac{2\left(1+\frac{2}{m}\right)}{\left(1+\frac{1}{m}\right)(1+\frac 3m)}\right|\end{aligned}$
Now
$\lim \left|\frac{a_{m+1}}{a_{m}}\right|=2 \in (0,\infty)$
Then, The given power series is absolutely convergent i.e., convergent $\forall x$ such that $|x+2|<\frac 12$
Then the answer is $\frac 12$
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