If alpha, beta are the roots of the equation, | vedclass

Name: PDF Generator App | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

${x}^2-\mathrm{x}-1=0 $

$\mathrm{~S}_{\mathrm{n}}=2023 \alpha^{\mathrm{n}}+2024 \beta^{\mathrm{n}} $

$ \mathrm{S}_{\mathrm{n}-1}+\mathrm{S}_{\ma

Vedclass · Accepted Answer

$\mathrm{S}_{12}=\mathrm{S}_{11}+\mathrm{S}_{10}$

Vedclass · Answer

${x}^2-\mathrm{x}-1=0 $

$\mathrm{~S}_{\mathrm{n}}=2023 \alpha^{\mathrm{n}}+2024 \beta^{\mathrm{n}} $

$ \mathrm{S}_{\mathrm{n}-1}+\mathrm{S}_{\mathrm{n}-2}=2023 \alpha^{\mathrm{n}=1}+2024 \beta^{\mathrm{n}-1}+2023 \alpha^{\mathrm{n}-2}+2024 \beta^{\mathrm{n}-2} $

$ =2023 \alpha^{\mathrm{n}-2}[1+\alpha]+2024 \beta^{\mathrm{n}-2}[1+\beta] $

$=2023 \alpha^{\mathrm{n}-2}\left[\alpha^2\right]+2024 \beta^{\mathrm{n}-2}\left[\beta^2\right] $

$ =2023 \alpha^{\mathrm{n}}+2024 \beta^{\mathrm{n}} $

$ \mathrm{S}_{\mathrm{n}-1}+\mathrm{S}_{\mathrm{n}-2}=\mathrm{S}_{\mathrm{n}} $

$ \mathrm{Put} \mathrm{n}=12 $

$\mathrm{~S}_{11}+\mathrm{S}_{10}=\mathrm{S}_{12}$

If $\alpha, \beta$ are the roots of the equation, $x^2-x-1=0$ and $S_n=2023 \alpha^n+2024 \beta^n$, then

Similar Questions

The equation $\sqrt {3 {x^2} + x + 5} = x - 3$ , where $x$ is real, has

Let $\alpha, \beta$ be two roots of the equation $x^{2}+(20)^{\frac{1}{4}} x+(5)^{\frac{1}{2}}=0$. Then $\alpha^{8}+\beta^{8}$ is equal to:

Equation $\frac{3}{{x - {a^3}}} + \frac{5}{{x - {a^5}}} + \frac{7}{{x - {a^7}}} = 0,a > 1$ has

If the equation $\frac{{{x^2} + 5}}{2} = x - 2\cos \left( {ax + b} \right)$ has atleast one solution, then $(b + a)$ can be equal to

Let $\alpha$ and $\beta$ be the roots of the equation $5 x^{2}+6 x-2=0 .$ If $S_{n}=\alpha^{n}+\beta^{n}, n=1,2,3 \ldots$ then :