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?

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## Comments

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

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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?

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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,

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

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?

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$a$ in terms of $x$, $b$ in terms of $y$ and substitute in $S$?

Oh okay. But isn't that the same as what I did? FindLog in to reply

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