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?

#Sequences #Series #SumOfSeries #SumOfSequence #SummationOfSeries

Easy Math Editor

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`2 \times 3`	$2 \times 3$
`2^{34}`	$2^{34}$
`a_{i-1}`	$a_{i-1}$
`\frac{2}{3}`	$\frac{2}{3}$
`\sqrt{2}`	$\sqrt{2}$
`\sum_{i=1}^3`	$\sum_{i=1}^3$
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Comments

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

Pi Han Goh - 5 years, 11 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?

Omkar Kulkarni - 5 years, 11 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,

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

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?

Pi Han Goh - 5 years, 11 months ago

@Pi Han Goh – 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 $S$ ?

Omkar Kulkarni - 5 years, 11 months ago

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

Pi Han Goh - 5 years, 11 months ago

@Pi Han Goh – Right. Thanks! :D

Omkar Kulkarni - 5 years, 11 months ago

Math	Appears as
Remember to wrap math in `\(` ... `\)` or `\[` ... `\]` to ensure proper formatting.
`2 \times 3`	$2 \times 3$
`2^{34}`	$2^{34}$
`a_{i-1}`	$a_{i-1}$
`\frac{2}{3}`	$\frac{2}{3}$
`\sqrt{2}`	$\sqrt{2}$
`\sum_{i=1}^3`	$\sum_{i=1}^3$
`\sin \theta`	$\sin \theta$
`\boxed{123}`	$\boxed{123}$