Question

Help me to prove this!!!

Answer 1

If the roots of the quadratic expression
\[x ^{2} -px -q =0\] are \[\alpha \ and \ \beta \]

then prove
\[\alpha ^{n} + \beta ^{n} = p[\alpha ^{n-1} + \beta ^{n-1}] + q[\alpha ^{n-2} + \beta ^{n-2}]\]

Answer 2

@ganeshie8 , @mathstudent55 , @thomaster , @phi help me

Answer 3

Well, you know that \((x-\alpha)(x-\beta)=x^2-(\alpha+\beta)+\alpha\beta=x^2-px-q\). Thus, \(p=\alpha+\beta\), and \(q=-\alpha\beta\). Plug those values for \(p,q\) in the equation\[p[\alpha ^{n-1} + \beta ^{n-1}] + q[\alpha ^{n-2} + \beta ^{n-2}]\] and simplify.

Answer 4

Sorry.... I know that !!! But I have to Prove it other way around...:/

Answer 5

What do you mean the "other way around?" Since you just have a bunch of equalities you can completely reverse the argument if that's what you want.

Answer 6

no!
the question asks to build a expression for
\[\alpha ^{n} + \beta ^{n} \]
and All I know is that the answer is wt I got on left side...... I don't know how to deduce it

Answer 7

\[\begin{aligned}
\alpha^n+\beta^n&=\alpha^n+\beta^n+\alpha\beta^{n-1}-\alpha\beta^{n-1}+\beta\alpha^{n-1}-\beta\alpha^{n-1}\\
&=\alpha^n+\alpha\beta^{n-1}+\beta\alpha^{n-1}+\beta^n-\alpha\beta^{n-1}-\beta\alpha^{n-1}\\
&=(\alpha+\beta)(\alpha^{n-1}+\beta^{n-1})-\alpha\beta(\alpha^{n-2}+\beta^{n-2})\\
&=p[\alpha ^{n-1} + \beta ^{n-1}] + q[\alpha ^{n-2} + \beta ^{n-2}]
\end{aligned}\]You just reverse the obvious way of proving it's true. There's no simpler way to prove this.

Answer 8

Unless a specific *method* has been described, go with this.