# How to proceed further in this Arithmetico-Geometric Progression problem

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.

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 $$n(4n-3)$$

• Nice work; it has to be an infinite series Commented Aug 15, 2019 at 13:04
• $\cdots$ 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 Commented Aug 15, 2019 at 13:09
• @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.!!!! Commented Aug 15, 2019 at 13:25
• 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... Commented Aug 15, 2019 at 13:33
• @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. Commented Aug 15, 2019 at 13:47

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

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 $$s_n=n(4n-3)$$

• Thank you for your answer sir. Commented Aug 15, 2019 at 13:26
• Thanks for your attention Commented Aug 15, 2019 at 13:35

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

• Thank you for your answer sir. Commented Aug 15, 2019 at 13:26
• This was only your work! Commented Aug 15, 2019 at 13:27
• Yes sir. I didn't know how to proceed which I learnt in the accepted answer. I thanked you for spending your time in trying to clarifying my doubt. Commented Aug 15, 2019 at 13:28
• It was just a simple substitution which would have made the steps easier. Commented Aug 15, 2019 at 13:29