Solveeit Logo
Login

Question

Mathematics Question on Quadratic Equations

If α\alpha and β\beta are the roots of x2−ax+b=0x^{2}-ax+b=0 and if αn+βn=Vn,\alpha^{n}+\beta^{n}=V_{_n}, then

A

Vn+1=aVn+bVn−1V_{_{n+1}}= aV_{_n}+ bV_{_{n-1}}

B

Vn+1=aVn+aVn−1V_{_{n+1}}= aV_{_n}+ aV_{_{n-1}}

C

Vn+1=aVn−bVn−1V_{_{n+1}}= aV_{_n}- bV_{_{n-1}}

D

Vn+1=aVn−1−bVnV_{_{n+1}}= aV_{_n-1}- bV_{_{n}}

Answer

Vn+1=aVn−bVn−1V_{_{n+1}}= aV_{_n}- bV_{_{n-1}}

Explanation

Solution

Multiplying x2−ax+b=0x^{2}-a x+b=0 by xn−1x^{n-1} xn+1=axn+bxn−1=0…x^{n+1}=a x^{n}+b x^{n-1}=0\ldots(i) α,β\alpha, \beta are roots of x2−ax+b=0x^{2}-a x+b=0, therefore they will satisfy (i). Also, αn+1−aαn+bαn−1=0…\alpha^{n+1}-a \alpha^{n}+b \alpha^{n-1}=0 \ldots(ii) and βn+1−aβn+bβn−1=0\beta^{n+1}-a \beta^{n}+b \beta^{n-1}=0 On adding Eqs. (ii) and (iii), we get (αn+1+βn+1)−a(αn+βn)\left(\alpha^{n+1}+\beta^{n+1}\right)-a\left(\alpha^{n}+\beta^{n}\right) +b(αn−1+β(n−1))=0+b\left(\alpha^{n-1}+\beta^{(n-1)}\right)=0 or Vn+1−aVn+bVn−1=0V_{n+1}-a V_{n}+b V_{n-1}=0 ⇒Vn+1=aVn−bVn−1\Rightarrow V_{n+1}=a V_{n}-b V_{n-1} (given, αn+βn=Vn)\left.\alpha^{n}+\beta^{n}=V_{n}\right)