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# Doubt!

Let $x=1+3a+6a^2+10a^3+\dots$ and $y=1+4b+10b^2+20b^3+\dots$ for $$\lvert a\rvert<1$$ and $$\lvert b\rvert<1$$. Find $S=1+3(ab)+5(ab)^2+\dots$ in terms of $$x$$ and $$y$$.

I found $$x$$ and $$y$$ in terms of $$a$$ and $$b$$ respectively, and then substituted in a closed expression for $$S$$, which came out to be pretty complicated. This leads me to think that I'm wrong somewhere. Also, is there a method faster than finding closed expressions for $$x$$ and $$y$$ and then substituting or whatever?

Note by Omkar Kulkarni
2 years, 3 months ago

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Hint: Differentiate the series $$\displaystyle \sum_{n=1}^\infty a^n = \frac a{1-a}$$ with respect to $$a$$.

- 2 years, 3 months ago

Nope, not getting it. Also, we haven't learned differentiation yet, so I can't say I'm familiar enough with derivatives. Could you explain?

- 2 years, 3 months ago

Okay, I will do for $$x$$. I take it that $$1,3,6,10,\ldots$$ are all triangular numbers. then $$x = \displaystyle \sum_{n=1}^\infty \frac {n(n+1)}2 a^{n-1}$$.

If we differentiate $$\displaystyle \sum_{n=1}^\infty a^n = \frac a{1-a}$$, we get $$\displaystyle \sum_{n=1}^\infty na^{n-1} = \frac 1{(1-a)^2 }$$. Differentiate once more, we get $$\displaystyle \sum_{n=1}^\infty n(n-1)a^{n-2} = \frac 2{(1-a)^3 }$$. Then,

$\begin{eqnarray} \displaystyle 2x &=& \sum_{n=1}^\infty n(n+1) a^{n-1} \\ &=& \sum_{n=1}^\infty (n(n-1) + 2n) a^{n-1} \\ &=& \sum_{n=1}^\infty n(n-1) a^{n-1} + 2 \sum_{n=1}^\infty n a^{n-1} \\ &=& a \sum_{n=1}^\infty n(n-1) a^{n-2} + 2 \sum_{n=1}^\infty n a^{n-1} \\ &=& a \cdot \frac2{(1-a)^3} + 2 \cdot \frac1{(1-a)^2} \\ &=& \frac{2a}{(1-a)^3} \\ \end{eqnarray}$

So $$x = \frac{a}{(1-a)^3}$$. Using the same concept, we can state $$y$$ in terms of $$b$$.

Now, a few tricks for $$y$$: I assume $$1,4,10,20,\ldots$$ are tetratehedral numbers, that is $$T_n = \frac{n(n+1)(n+2)} 6$$. Your mission here is to state $$n(n+1)(n+2)$$ in terms of a linear combination of $$n(n-1)(n-2) , n(n-1)$$ and $$n$$ like what I've done above for $$x$$.

Another thing to note that $$S$$ is an arithmetic geometric progression.

Can you finish it off?

- 2 years, 3 months ago

Oh okay. But isn't that the same as what I did? Find $$a$$ in terms of $$x$$, $$b$$ in terms of $$y$$ and substitute in $$S$$?

- 2 years, 3 months ago

I don't know what method you did, I just input my thoughts on how to solve it.

- 2 years, 3 months ago

Right. Thanks! :D

- 2 years, 3 months ago