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# Ramanujan's Polynomial.

$7x^2 + 5x + 1729$

The quadratic polynomial mentioned above has roots $$\alpha$$ and $$\beta$$.

Let , $$S_n = \alpha^{n-1} +\beta^{n-1}$$

Then evaluate :

$\sum_{n=1}^{\infty} (S_n)^2$

7 months ago

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Comment deleted 7 months ago

Use Infinite GP formula after this. Easy. · 7 months ago

Is it required to prove the convergency of the summation then ? · 7 months ago

Comment deleted 7 months ago

@Chinmay Sangawadekar What have you tried?

The question seems easy. I have the answer in terms of variables, although I didn't calculate (I'm too lazy :P) · 7 months ago

Indeed it is easy...:P · 7 months ago

Did you use of infinite GP summation formula ? · 7 months ago

Yes I did.

You have $\displaystyle\sum_{n=1}^{\infty}S_n^2\\= \displaystyle\sum_{n=1}^{\infty}\alpha^{2n-2} +\beta^{2n-2}+2(\alpha\beta)^{n-1}$

Which is sum to infinity of $$\dfrac {a^{2n}}{a^2}$$ + Sum to infinity of $$\dfrac {b^{2n}}{b^2}$$ +sum to infinity of $$\dfrac {2(ab)^n}{ab}$$

Now, in these summations, $$\dfrac {1}{a^2}, \dfrac {1}{b^2}, \dfrac {2}{ab}$$ are constants. so we take them "out" of the summation, and then use the infinite gp summation formula.

Which leads to: $$\dfrac {2-(a^2+b^2)}{1-(a^2+b^2) +(ab)^2}+ \dfrac {2}{1-ab}$$

Use Vieta's and you're done.

Note, a= alpha

b=beta. · 7 months ago

First you can also show lim as n tends to infinity... Just to elabotrate... · 7 months ago

Exactly ! · 7 months ago

What caused problems earlier? · 7 months ago