If a_(1)=2,a_(2)=0 and a_(n)=3a_(n-1)-3a_(n-2) then find the value of a_(5). Answer:

Initialize Values: To find the value of a_(5), we need to use the recursive formula a_(n)=3a_(n-1)-3a_(n-2) and the initial values a_(1)=2 and a_(2)=0. We will calculate the values of a_(3), a_(4), and then a_(5) in sequence.Calculate a_(3): First, let's find a_(3) using the recursive formula. We have a_(3)=3a_(2)-3a_(1). Substituting the known values, we get a_(3)=3*0-3*2.Calculate a_(4): Calculating a_(3), we get a_(3)=0-6=-6.Calculate a_(5): Next, we find a_(4) using the recursive formula. We have a_(4)=3a_(3)-3a_(2). Substituting the known values, we get a_(4)=3*(-6)-3*0.Final Result: Calculating a_(4), we get a_(4)=-18-0=-18.Finally, we find a_(5) using the recursive formula. We have a_(5)=3a_(4)-3a_(3). Substituting the known values, we get a_(5)=3*(-18)-3*(-6).Calculating a_(5), we get a_(5)=-54+18=-36.

Home

Math Problems

Precalculus

Find the roots of factored polynomials

Full solution

Q. If

a_{1}=2, a_{2}=0

and

a_{n}=3 a_{n-1}-3 a_{n-2}

then find the value of

a_{5}

\newline

Answer:

Initialize Values: To find the value of $a_{5}$ , we need to use the recursive formula $a_{n}=3a_{n-1}-3a_{n-2}$ and the initial values $a_{1}=2$ and $a_{2}=0$ . We will calculate the values of $a_{3}$ , $a_{4}$ , and then $a_{5}$ in sequence.
Calculate $a_{3}$ : First, let's find $a_{3}$ using the recursive formula. We have $a_{3}=3a_{2}-3a_{1}$ . Substituting the known values, we get $a_{3}=3\times 0-3\times 2$ .
Calculate $a_{4}$ : Calculating $a_{3}$ , we get $a_{3}=0-6=-6$ .
Calculate $a_{5}$ : Next, we find $a_{4}$ using the recursive formula. We have $a_{4}=3a_{3}-3a_{2}$ . Substituting the known values, we get $a_{$4$}=$3$*($-6$)$-3$*$0$.
Final Result: Calculating $a_{4}$, we get $a_{4}=-18-0=-18$.
Final Result: Calculating $a_{4}$, we get $a_{4}=-18-0=-18$. Finally, we find $a_{5}$ using the recursive formula. We have $a_{5}=3a_{4}-3a_{3}$. Substituting the known values, we get $a_{5}=3*(-18)-3*(-6)$.
Final Result: Calculating $a_{4}$, we get $a_{4}=-18-0=-18$. Finally, we find $a_{5}$ using the recursive formula. We have $a_{5}=3a_{4}-3a_{3}$. Substituting the known values, we get $a_{5}=3*(-18)-3*(-6)$. Calculating $a_{5}$, we get $a_{5}=-54+18=-36$.