MHB Finding $S_{2013}^2$ from a Given Sequence

  • Thread starter Thread starter Albert1
  • Start date Start date
  • Tags Tags
    Sequence
Albert1
Messages
1,221
Reaction score
0
a sequence ${a_n>0}$ for all n

given :$S_n - \dfrac{1}{a_n}=a_n-S_n$

find :$S_{2013}^2$

where :$S_n=a_1+a_2+-------+a_n$
 
Mathematics news on Phys.org
One can easily find the first terms in the series:
\[a_1=\sqrt{1}-\sqrt{0} \\ a_2 = \sqrt{2}-\sqrt{1} \\ a_3 = \sqrt{3}-\sqrt{2}\]
Suppose this system holds for the nth level (n > 3): $a_n = \sqrt{n}-\sqrt{n-1}$ and let´s find $a_{n+1}$:
\[2S_{n+1} = a_{n+1}+\frac{1}{a_{n+1}} \\\\ 2(\sqrt{n}+a_{n+1}) = a_{n+1}+\frac{1}{a_{n+1}} \\\\ a_{n+1}^2 + 2\sqrt{n}\cdot a_{n+1}-1 = 0 \\\\ a_{n+1} = \frac{1}{2}\left ( -2\sqrt{n}\; \pm \sqrt{4n+4} \right ) \\\\ a_{n+1} = \sqrt{n+1}-\sqrt{n}\]
By the principle of induction $a_n=\sqrt{n}-\sqrt{n-1}$ and $S_n=\sqrt{n}$ for $n = 1,2,3,..$
Thus $S_{2013}^2 = (\sqrt{2013})^2 = 2013$
 
lfdahl said:
One can easily find the first terms in the series:
\[a_1=\sqrt{1}-\sqrt{0} \\ a_2 = \sqrt{2}-\sqrt{1} \\ a_3 = \sqrt{3}-\sqrt{2}\]
Suppose this system holds for the nth level (n > 3): $a_n = \sqrt{n}-\sqrt{n-1}$ and let´s find $a_{n+1}$:
\[2S_{n+1} = a_{n+1}+\frac{1}{a_{n+1}} \\\\ 2(\sqrt{n}+a_{n+1}) = a_{n+1}+\frac{1}{a_{n+1}} \\\\ a_{n+1}^2 + 2\sqrt{n}\cdot a_{n+1}-1 = 0 \\\\ a_{n+1} = \frac{1}{2}\left ( -2\sqrt{n}\; \pm \sqrt{4n+4} \right ) \\\\ a_{n+1} = \sqrt{n+1}-\sqrt{n}\]
By the principle of induction $a_n=\sqrt{n}-\sqrt{n-1}$ and $S_n=\sqrt{n}$ for $n = 1,2,3,..$
Thus $S_{2013}^2 = (\sqrt{2013})^2 = 2013$
very good solution :)
 
Insights auto threads is broken atm, so I'm manually creating these for new Insight articles. In Dirac’s Principles of Quantum Mechanics published in 1930 he introduced a “convenient notation” he referred to as a “delta function” which he treated as a continuum analog to the discrete Kronecker delta. The Kronecker delta is simply the indexed components of the identity operator in matrix algebra Source: https://www.physicsforums.com/insights/what-exactly-is-diracs-delta-function/ by...
Suppose ,instead of the usual x,y coordinate system with an I basis vector along the x -axis and a corresponding j basis vector along the y-axis we instead have a different pair of basis vectors ,call them e and f along their respective axes. I have seen that this is an important subject in maths My question is what physical applications does such a model apply to? I am asking here because I have devoted quite a lot of time in the past to understanding convectors and the dual...
Back
Top