0
$\begingroup$

Need help! I'm computing for the general formula of this sequence $S = 1 + 4t^2 + 9t^4 + \ldots + n^2 t^{2n - 2}$. I tried multiplying the equation by $t^2$ then subtracting it by the original equation..I do this twice... but I only got few terms correct..

the correct answer based on an online calculator is $n^2 t^{2n} – 2n^2 t^{2n+2} + n^2 t^{2n+4} + 2nt^{2n} + t^{2n} – 2nt^{2n+2} + t^{2n+2} – t^2 - 1$.

How is this possible? I got $7$ terms correct...but nowhere to find the other $2$..I check my work over and over...but still nothing come up...

$\endgroup$
1
  • $\begingroup$ You did notice that you can collect terms in that result right? $\endgroup$
    – WimC
    Apr 27, 2015 at 6:25

2 Answers 2

2
$\begingroup$

You're trying to find $$S_n(t) = \sum_{j=1}^n j^2 t^{2j-2}$$ Hint: start with the geometric series $$ A_n(t) = \sum_{j=1}^n t^{2j}$$ and differentiate twice.

$\endgroup$
1
  • $\begingroup$ You can also use a modification of "differentiate twice" with $\frac 1t(tA'_n)'$. Multiplying by $t$ before differentiating the second time helps with the coefficient. $\endgroup$ Apr 27, 2015 at 7:39
0
$\begingroup$

Your method can indeed be used. Perhaps you made an accidental mistake? Also, you missed a $\dfrac{1}{(t^2-1)^3}$ in the expected answer.

Note that it works only when $t^2 \ne 1$.

$S (1-t^2) = T - n^2 t^{2n}$ where $T = 1 + 3 t^2 + 5 t^4 + ... + (2n-1) t^{2n-2}$.

$T (1-t^2) = U - (2n-1) t^{2n} - 1$ where $U = 2 + 2 t^2 + 2 t^4 + ... + 2 t^{2n-2}$

$U (1-t^2) = 2 (1-t^{2n})$.

Therefore:

$S = \dfrac{ n^2 t^{2n} - T }{ 1-t^2 }$

$= \dfrac{ n^2 t^{2n} }{ t^2-1 } - \dfrac{ (2n-1) t^{2n} + 1 - U }{ (t^2-1)^2 }$

$= \dfrac{ n^2 t^{2n} }{ t^2-1 } - \dfrac{ (2n-1) t^{2n} + 1 }{ (t^2-1)^2 } + \dfrac{ 2(t^{2n}-1) }{ (t^2-1)^3 }$.

$\endgroup$
2
  • $\begingroup$ @ user21820 tnx a lot sir :) I already got it right... but your solution is neat and clean as to compare to my work,, hehehe tnx a lot!! $\endgroup$
    – rosa
    May 2, 2015 at 1:50
  • 1
    $\begingroup$ @rosa: You're welcome! Notice that there is a more powerful technique using the forward difference operator. Basically you define $D(f) = ( n \mapsto f(n+1) - f(n) )$ for any function $f$ on integers. Then you just need to find an anti-difference of the general term in your summation, so that it becomes a telescoping sum. To do so you would need to use anti-difference by parts (derived in the same way as differentiation by parts) and the fact that $c^n$ has an easy anti-difference for any $c$. Doing this twice will yield the answer. $\endgroup$
    – user21820
    May 2, 2015 at 5:40

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .