Let x=1+3a+6a2+10a3+x=1+3a+6a^2+10a^3+\dots and y=1+4b+10b2+20b3+y=1+4b+10b^2+20b^3+\dots for a<1\lvert a\rvert<1 and b<1\lvert b\rvert<1. Find S=1+3(ab)+5(ab)2+S=1+3(ab)+5(ab)^2+\dots in terms of xx and yy.


I found xx and yy in terms of aa and bb respectively, and then substituted in a closed expression for SS, 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 xx and yy and then substituting or whatever?

Note by Omkar Kulkarni
4 years, 3 months ago

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Hint: Differentiate the series n=1an=a1a \displaystyle \sum_{n=1}^\infty a^n = \frac a{1-a} with respect to aa.

Pi Han Goh - 4 years, 3 months ago

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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?

Omkar Kulkarni - 4 years, 3 months ago

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Okay, I will do for xx. I take it that 1,3,6,10,1,3,6,10,\ldots are all triangular numbers. then x=n=1n(n+1)2an1 x = \displaystyle \sum_{n=1}^\infty \frac {n(n+1)}2 a^{n-1} .

If we differentiate n=1an=a1a \displaystyle \sum_{n=1}^\infty a^n = \frac a{1-a} , we get n=1nan1=1(1a)2 \displaystyle \sum_{n=1}^\infty na^{n-1} = \frac 1{(1-a)^2 } . Differentiate once more, we get n=1n(n1)an2=2(1a)3 \displaystyle \sum_{n=1}^\infty n(n-1)a^{n-2} = \frac 2{(1-a)^3 } . Then,

2x=n=1n(n+1)an1=n=1(n(n1)+2n)an1=n=1n(n1)an1+2n=1nan1=an=1n(n1)an2+2n=1nan1=a2(1a)3+21(1a)2=2a(1a)3 \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=a(1a)3 x = \frac{a}{(1-a)^3} . Using the same concept, we can state yy in terms of bb.

Now, a few tricks for yy: I assume 1,4,10,20,1,4,10,20,\ldots are tetratehedral numbers, that is Tn=n(n+1)(n+2)6T_n = \frac{n(n+1)(n+2)} 6 . Your mission here is to state n(n+1)(n+2)n(n+1)(n+2) in terms of a linear combination of n(n1)(n2),n(n1) n(n-1)(n-2) , n(n-1) and nn like what I've done above for xx.

Another thing to note that SS is an arithmetic geometric progression.

Can you finish it off?

Pi Han Goh - 4 years, 3 months ago

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@Pi Han Goh Oh okay. But isn't that the same as what I did? Find aa in terms of xx, bb in terms of yy and substitute in SS?

Omkar Kulkarni - 4 years, 3 months ago

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

Pi Han Goh - 4 years, 3 months ago

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@Pi Han Goh Right. Thanks! :D

Omkar Kulkarni - 4 years, 3 months ago

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