\[7x^2 + 5x + 1729\]

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

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

Then evaluate :

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

\[7x^2 + 5x + 1729\]

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

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

Then evaluate :

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

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– Mehul Arora · 10 months, 4 weeks ago

Use Infinite GP formula after this. Easy.Log in to reply

– Chinmay Sangawadekar · 10 months, 4 weeks ago

Is it required to prove the convergency of the summation then ?Log in to reply

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@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) – Mehul Arora · 10 months, 4 weeks ago

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– Chinmay Sangawadekar · 10 months, 4 weeks ago

Indeed it is easy...:PLog in to reply

– Chinmay Sangawadekar · 10 months, 4 weeks ago

Did you use of infinite GP summation formula ?Log in to reply

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. – Mehul Arora · 10 months, 4 weeks ago

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– Chinmay Sangawadekar · 10 months, 4 weeks ago

First you can also show lim as n tends to infinity... Just to elabotrate...Log in to reply

@Chinmay Sangawadekar – Mehul Arora · 10 months, 4 weeks ago

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– Chinmay Sangawadekar · 10 months, 4 weeks ago

Exactly !Log in to reply

– Mehul Arora · 10 months, 4 weeks ago

What caused problems earlier?Log in to reply

– Chinmay Sangawadekar · 10 months, 4 weeks ago

No problems ... why ?Log in to reply