Recursive Sequence

$a_n = 40 - 4a_{n-1} a_n - 8 = -4(a_{n-1} - 8)$ . Let $b_n = a_n - 8$ for all natural numbers $n$ . Then, the original equation becomes $b_n = -4(b_{n-1})$ , from which we conclude that $b_n = (-4)^n b_0$ . Since $b_0 = a_0 - 8 = -4 - 8 = -12$ , thus $b_n = a_n - 8 = (-12) (-4)^n$ . Therefore $a_i = (-12) (-4)^i + 8$ . We thus calculate that $r^2 + s^2 + t^2 = (-12)^2 + (-4)^2 + (8)^2 = 224$ .

[Latex Edits, edits for clarity - Calvin]

When $i = 0$ , $r + t = -4$ . ---(1)

When $i = 1$ , $r \cdot s + t = 40 - 4 \cdot (-4) = 56$ . ---(2)

When $i = 2$ , $r \cdot s^2 + t = 40 - 4 \cdot 56 = -184$ . ---(3)

(2)–(1) $r \cdot (s-1) = 60$ .

(3)–(2) $r \cdot s \cdot (s-1) = -240$ .

If we divide the 2 equations, we get $s = -4$ . Then $r = \frac {60}{-5} = -12$ . Substituting r into (1), we get $t= 8$ .

So $r^2 + s^2 + t^2 = (-12)^2 + (-4)^2 + 8^2 = 224$

One has r x s^2 + t = - 184, r x s + t = 56, r + t = - 4. Clearly, -t = r + 4. One has s = (56 - t) / r = 60 / r + 1. One has s^2 = (-184 - t) / r = - 180 / r + 1. Now, s^2 + 3 s = - 180/r + 1 + 180/r + 3, therefore s^2 + 3 x s = 4, and s 1 = 1, s 2 = -4. But s<>1, because - 184 = r + t <> 56 = r + t. Then s = - 4. One obtain -4 x r + t = 56, r + t = - 4, so 5 x r = -60, and r = - 12 and t = 8. Consequently, r^2 + s^2 + t^2 = (-12)^2 + (-4)^2 + 8^2 = 144 + 16 + 64 = 144 + 80 = 224.

by using the recurrence relation we obtain a 1 = rs + t = 56, a 2 = rs^2 + t = -184, a_3 = rs^3 + t = 776. Simply by rearranging those equations we can get s = \frac {776 - t}{-184-t} = \frac{-184-t}{56-t}. Solving this we get t=8 and by plugging back t=8 we get s=-4. Lastly, 56 = r \cdot (-2) + 8, giving us r = -12. So r^2 +s^2 +t^2 = 144 + 16 + 64 = 224.

I started by listing out the first few terms of a: -4, 56, -184 Next I equated them in terms of r, s and t: r + t = -4 rs + t = 56 rs^2 + t = -184 Then, I subtracted adjacent terms and got: rs - r = 60 rs^2 - rs = s(rs - r) = s(60) = -240 By factoring s out of the second equation and plugging in 60 for rs - r we can get s = -4. Finally we can solve for r and t by plugging in s back in to find r and then r back in to find t. And we get r = -12 and t = 8.

$a_0$ = -4 $a_1$ = 56 $a_2$ = -184 r + t = -4
rs + t = 56 r $s^2$ + t = -184

By solving these equations,

we get s = -4 r = -12 t = 8

$r^2$ + $s^2$ + $t^2$ = $(-4)^2$ + $(-12)^2$ + $8^2$

$r^2$ + $s^2$ + $t^2$ = 224

As an = 40- 4an-1 and ai = r • si + t, a0 = -4 = r + t; ----------- 1 a1 = 56 = rs + t -----------2 a2 = -184 = rs2 + t ----------3 2–1: rs – r = r(s-1) = 60 ----------4 3-1: rs2 – r = -180 ----------5 3-2: rs2 – rs = rs(s-1) = -240 ----------6 6-4: (s-1)(rs-r) = -300 ----------7 7÷4: s-1 = -5 s=-4 4: r(s-1) = 60 -5r = 60 r = -12 1: r + t = -4 t= 8 Therefore, r2 + s2 + t2 = 144 + 16 + 64 = 224

a(0) = -4 = r+t a(1) = 40-4(-4) = 56 = rs+t a(2) = 40--(-56) = -184 = rs^2 +t

From a(0) and a(1) we can get 60 = r(s-1)

From a(0) and a(2) we can get -180 = r(s-1)(s+1)

From these two equations we can get r = -12 , s = -4 , t = 8

r^2 + s^2 + t^2 = 224

put i=0,1,2 for n in an=40−4an−1 ai=r⋅si+t then we get r+t=-4 rs+t=56 rs.s+t=-184 solving the above eq we get r=-12,s=-4,t=8 put in the result we get r2+s2+t2=224

eq1: r+t=-4 eq2: rs+t=56 eq3: r(s^2)+t=-184

eq2 - eq1: r(s-1)=60 eq3 - eq2: rs(s-1)=-240 therefore, s=-4 it follows that r=-12 and t=8

r^2+s^2+t^2=(-12)^2+(-4)^2+8^2=224

Solution 1: Set $b_n = a_n - 8$ . Then, $b_0 = - 12$ . We also have that $b_n = 40 - 4a_{n-1} - 8 = 32 - 4a_{n-1} = -4(a_{n-1} - 8) = -4b_{n-1}$ . Substituting recursively, we have, $b_n = (-4)\times b_{n-1} = (-4)^2\times b_{n-2} = (-4)^{n} \times b_0 = (-12)(-4)^n$ . Thus, $a_n = b_n + 8 = (-12)(-4)^n + 8$ , for all non-negative $n$ , with $r=-12, s = -4$ and $t = 8$ . Therefore, $r^2+s^2+t^2 = 144 + 16 + 64 = 224$ .

Solution 2: Substitute the condition $a_i = r\cdot s^i + t$ into $a_n = 40 - 4a_{n-1}$ . We get, $r\cdot s^n + t = 40 - 4r\cdot s^{n-1} - 4t$ . Rearranging, we have, $(rs+4r) \cdot s^{n-1} =40- 5t$ . Since the RHS is constant over all $n$ , the LHS must be constant over all $n$ . If $s^{n-1}$ is a constant (i.e. $s=0$ or $1$ ), then $a_i$ will be a constant sequence, which is a contradiction. Hence, $s^{n-1}$ is not a constant, so the coefficient $(rs+4r)$ must be the constant 0. Likewise, if $r=0$ , then $a_i$ will be a constant sequence, hence $r\neq 0$ and $s+4 = 0 \Rightarrow s=-4$ . This gives us that $40-5t=0 \Rightarrow t=8$ . Now, subsituting in the value at $i=0$ , we get $-4 = r (-4)^0 + 8 \Rightarrow r = -12$ . Hence, $a_i = (-12)(-4)^i + 8$ , and $r^2+s^2+t^2=224$ .

Three unknowns, therefore, generate three equations to solve.

Solve them any way you like.