MHB Series Simplification: Simplifying \sum^{\infty}_{n = 1} 4^n/3^{n-1}

  • Thread starter Thread starter shamieh
  • Start date Start date
  • Tags Tags
    Series
AI Thread Summary
The discussion focuses on simplifying the infinite series \(\sum^{\infty}_{n = 1} \frac{4^n}{3^{n-1}}\) into the form \(ar^{n-1}\). The transformation involves recognizing that \(4^n\) can be expressed as \(4^{1+(n-1)}\), which simplifies to \(4 \cdot 4^{n-1}\). This allows the series to be rewritten as \(\sum^{\infty}_{n = 1} 4\left(\frac{4}{3}\right)^{n-1}\). The confusion arises from understanding where the extra factor of 4 comes from and how the exponent manipulation works. Ultimately, the algebraic steps clarify the series' conversion into the desired format.
shamieh
Messages
538
Reaction score
0
They are re writing it to be in the form $$ar^{n-1}$$ But I can't figure out how

$$\sum^{\infty}_{n = 1} \frac{4^n}{3^{n- 1}}$$ turns into this: $$\sum^{\infty}_{n = 1} 4(\frac{4}{3})^{n-1}$$ ??

I'm confused on the Algebra they are implementing to do this..Can anyone elaborate? For example, where is the top n, where is the extra 4 coming from.. I know this is simple algebra but It's flying right over me :confused:
 
Mathematics news on Phys.org
shamieh said:
They are re writing it to be in the form $$ar^{n-1}$$ But I can't figure out how

$$\sum^{\infty}_{n = 1} \frac{4^n}{3^{n- 1}}$$ turns into this: $$\sum^{\infty}_{n = 1} 4(\frac{4}{3})^{n-1}$$ ??

I'm confused on the Algebra they are implementing to do this..Can anyone elaborate? For example, where is the top n, where is the extra 4 coming from.. I know this is simple algebra but It's flying right over me :confused:

From the laws of exponents you can say $$a^{b+c} = a^b\, a^c$$

In this case you can write $$4^n = 4^{1+(n-1)} = 4^1 \cdot 4^{n-1}$$
 
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...
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...

Similar threads

Back
Top