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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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,

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

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?

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