another fibonacci proof

**padsinseven** · October 15th 2007, 11:49 PM

Prove by induction that for all n>=1, u(n+1)^2 - u(n+2)u(n)= (-1)^n. Any help would be greatly appreciated. I have no idea how to show the right side will end up (-1)^n+1

**Soroban** · October 16th 2007, 07:10 AM

Hello, padsinseven!

This is the Fibonacci sequence: . $u_n\:=\:1,1,2,3,5,8,\cdots$

Prove by induction that for all $n \geq 1\!:\;\;\left(u_{n+1}\right)^2 - \left(u_{n+2}\right)\left(u_n\right)\;= \;(-1)^n$

Verify $S(1)\!:\;\;(u_2)^2 - (u_3)(u_1) \;=\;1^2 - (2)(1) \;=\;-1\;=\;(-1)^1$ . . . True!

Assume $S(k)$ is true: . $\left(u_{k+1}\right)^2 - \left(u_{k+2}\right)\left(u_k\right) \;=\;(-1)^k$

And we want prove $S(k+1)\!:\;\;\left(u_{k+2}\right)^2 - \left(u_{k+3}\right)\left(u_{k+1}\right) \;=\;(-1)^{k+1}$

The left side of $S(k+1)$ is:

. . . . . . $\left(u_{k+2}\right)^2 - \underbrace{\left(u_{k+3}\right)}\left(u_{k+1}\rig ht)$

. . $= \;\left(u_{k+2}\right)^2 - \overbrace{\left[u_{k+2} + u_{k+1}\right]}\left(u_{k+1}\right)$

. . $= \;\left(u_{k+2}\right)^2 - \left(u_{k+2}\right)\left(u_{k+1}\right) - \left(u_{k+1}\right)^2$

. . $= \;\left(u_{k+2}\right)\cdot\underbrace{\left[u_{k+2} - u_{k+1}\right]} - \left(u_{k+1}\right)^2$

. . $= \qquad\:\left(u_{k+2}\right)\cdot\overbrace{\left( u_k\right)} - \left(u_{k+1}\right)^2$

. . $= \;\qquad\; - \underbrace{\bigg[\left(u_{k+1}\right)^2 - \left(u_{k+2}\right)\left(u_k\right)\bigg]}_{\text{This is }S(k)}$

. . $= \;\qquad\qquad\qquad - \overbrace{(-1)^k}$

. . $= \qquad\qquad\qquad\quad (-1)^{k+1}$ . . . . ta-DAA!

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