The expression for a_n which satisfies a_0=0, | vedclass

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

Question

(A) Given the recurrence relation $a_n = a_{n-1} + a_{n-2}$ with $a_0 = 0$ and $a_1 = 1$. This is the characteristic equation $r^2 - r - 1 = 0$. The r

Vedclass · Accepted Answer

$\frac{1}{\sqrt{5}}\left(\frac{1+\sqrt{5}}{2}\right)^n-\frac{1}{\sqrt{5}}\left(\frac{1-\sqrt{5}}{2}\right)^n$

Vedclass · Answer

(A) Given the recurrence relation $a_n = a_{n-1} + a_{n-2}$ with $a_0 = 0$ and $a_1 = 1$. This is the characteristic equation $r^2 - r - 1 = 0$. The roots are $r = \frac{1 \pm \sqrt{5}}{2}$. Let $a_n = A\left(\frac{1+\sqrt{5}}{2}\right)^n + B\left(\frac{1-\sqrt{5}}{2}\right)^n$. Using $a_0 = 0$,we get $A + B = 0$,so $B = -A$. Using $a_1 = 1$,we get $A\left(\frac{1+\sqrt{5}}{2}\right) - A\left(\frac{1-\sqrt{5}}{2}\right) = 1$. $A\left(\frac{1+\sqrt{5}-1+\sqrt{5}}{2}\right) = 1$ $\Rightarrow A\sqrt{5} = 1$ $\Rightarrow A = \frac{1}{\sqrt{5}}$. Thus,$B = -\frac{1}{\sqrt{5}}$. Substituting these values,$a_n = \frac{1}{\sqrt{5}}\left(\frac{1+\sqrt{5}}{2}\right)^n - \frac{1}{\sqrt{5}}\left(\frac{1-\sqrt{5}}{2}\right)^n$.

The expression for $a_n$ which satisfies $a_0=0, a_1=1$ and $a_n=a_{n-1}+a_{n-2}, \forall n \in N -\{0,1\}$ is:

Similar Questions

$\sum_{k=1}^{\infty} \sum_{r=0}^k \frac{1}{3^k} \binom{k}{r}$ is equal to

What is the sum of $n$ terms of the series $2 + 5 + 14 + 41 + \dots$?

What is the $20^{\text{th}}$ term of the sequence defined by $a_{n} = (n-1)(2-n)(3+n)$?

$1+(1+3)+(1+3+5)+(1+3+5+7)+\ldots$ to $10$ terms $=$

Let $S_n = \sum_{k=1}^{4n} (-1)^{\frac{k(k+1)}{2}} k^2$. Then $S_n$ can take the value$(s)$:
$(A) 1056$
$(B) 1088$
$(C) 1120$
$(D) 1332$

Vedclass Test Series

Exam Paper Generator

Online Exam Module