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

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

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

Then evaluate :

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

No vote yet

1 vote

×

Problem Loading...

Note Loading...

Set Loading...

Easy Math Editor

`*italics*`

or`_italics_`

italics`**bold**`

or`__bold__`

boldNote: you must add a full line of space before and after lists for them to show up correctlyparagraph 1

paragraph 2

`[example link](https://brilliant.org)`

`> This is a quote`

Remember to wrap math in \( ... \) or \[ ... \] to ensure proper formatting.`2 \times 3`

`2^{34}`

`a_{i-1}`

`\frac{2}{3}`

`\sqrt{2}`

`\sum_{i=1}^3`

`\sin \theta`

`\boxed{123}`

## Comments

Sort by:

TopNewestComment deleted Feb 27, 2016

Log in to reply

Use Infinite GP formula after this. Easy.

Log in to reply

Is it required to prove the convergency of the summation then ?

Log in to reply

Comment deleted Feb 27, 2016

Log in to reply

@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)

Log in to reply

Indeed it is easy...:P

Log in to reply

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.

Log in to reply

Log in to reply

@Chinmay Sangawadekar

Log in to reply

Comment deleted 6 months ago

Log in to reply

Log in to reply

Log in to reply