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Question:

The sum to $n$ terms of the series,

$S=1+5(\frac{4n+1}{4n-3})+9(\frac{4n+1}{4n-3})^2+13(\frac{4n+1}{4n-3})^3+....$

The following image is my approach towards the problem.

Approach

Could you please tell how to proceed? I proceeded in this way as I did not want to mug-up the complicated formula given for AGP summation.

The final answer is $n(4n-3)$

Community
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Vishnu
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    Nice work; it has to be an infinite series – AgentS Aug 15 '19 at 13:04
  • @ganeshie8, Thank you. But in my book it is given to be sum upto n terms only. – Vishnu Aug 15 '19 at 13:06
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    $\cdots$ usually means going to infinity ;) and don't take your textbook too seriously - the author is using the same letter $n$ in the ratio and to count the terms – AgentS Aug 15 '19 at 13:09
  • @ganeshie8 Are you sure that isn't what he means? – saulspatz Aug 15 '19 at 13:11
  • I'm pretty sure – AgentS Aug 15 '19 at 13:12
  • @ganeshie8, I considered the summation to infinity and the answer matches. Thank you for your help. But lets see if anyone comes up with finite summation case. – Vishnu Aug 15 '19 at 13:14
  • You must divide your equation by $1-x$, if all is correct. – Dr. Sonnhard Graubner Aug 15 '19 at 13:14
  • @Dr.SonnhardGraubner, If I did so, the expression on the RHS became too complex, so that I stopped with the previous step itself. – Vishnu Aug 15 '19 at 13:15
  • Intellex looks @RobertZ has it with the finite terms. ok not so sure anymore -.- – AgentS Aug 15 '19 at 13:17
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    @ganeshie8, Thanks for spending your time sir. Doubt got cleared by RobertZ. Interestingly, in this problem the value of summation is same whether it is finite or infinite. Need to think more on this stuff. This is what makes mathematics interesting.!!!! – Vishnu Aug 15 '19 at 13:25
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    Totally agreeXD for now $\dfrac{4n+1}{4n-3}$ looks like a magic ratio that makes any partial sum equal to $n(4n-3)$. Infinite sum seems to diverge... – AgentS Aug 15 '19 at 13:33
  • @ganeshie8 Yes. What do you mean by infinite sum seems to diverge? – Vishnu Aug 15 '19 at 13:41
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    @ganeshie8 : Of course it diverges, $\sum (1+4k)x^k$ is combination of geometric and binomial series which has radius of convergence $|x|<1$. But $\frac{4n+1}{4n-3}>1$, so it is outside of the region of convergence. – Lutz Lehmann Aug 15 '19 at 13:47
  • @LutzL ahh right in that sum notation using $k$ for the index it's easy to see the sum diverges haha. Ty:) – AgentS Aug 15 '19 at 14:13
  • @Intellex diverges means the infinite sum doesn't converge. For example $1+\frac{1}{2}+\frac{1}{4}\cdots$ converges to $2$. But $1+2+3+4+\cdots$ diverges to $\infty$. If it seems circular, see http://mathworld.wolfram.com/DivergentSeries.html – AgentS Aug 15 '19 at 14:14
  • @ganeshie8, Thank you for the link. – Vishnu Aug 15 '19 at 14:15

3 Answers3

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Your approach is correct! Now for $x=\frac{4n+1}{4n-3}$, we find that $$S_n=\frac{x^n((4n-3)x-(4n+1))+(3x-1)}{(x-1)^2}=\frac{0+(3x-1)}{(x-1)^2}=n(4n-3).$$

Robert Z
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Note that $$x=1+\frac {4}{4n-3}$$ so $$1-x=\frac {-4}{4n-3}$$

Substitute in your last line and you will get the $$s_n=n(4n-3)$$

Mohammad Riazi-Kermani
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By your work $$S=\frac{1}{1-x}-4\frac{x(x^{n+1}-1)}{(x-1)^2}-\frac{(4n-3)x^n}{1-x}$$

Dr. Sonnhard Graubner
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