1. ## Proof by Induction

Hey guys,

Ive always have trouble with Inductions no matter how many Ive attempted to do. Please hint/help me please thanks.

f1^2 + f2^2 + f3^2....fn^2 = fn * fn+1
Im so lost in this one, I cant even figure out the base case....

2nd problem
is a summation problem

summation of i = 1 to n
i^2 = n(n+1)(2n+1)/6

base case of 1 works = 1^2 = 1 and summation = 6/6 = 1

assume p(n) is true for n = k
= k(k+1)(2k+1)/6

n = k+1

k(k+1)(2k+1)/6 + k+1

= k(k+1)(2k+1)/6 + 6k+6 = 2k3+2k^3+7k+6...Im stuck here.

2. Hello, ff4930!

You left out a huge chuck of the first problem . . .

We have the Fibonacci Sequence: 1, 1, 2, 3, 5, 8, 13, ...
. . where: . $F_1 = 1,\;F_2 =1,$ and: . $F_n \:=\:F_{n-1} + F_{n-2}$

Each term is the sum of the preceding two terms.

Prove: . $F_1^2 + F_2^2 + F_3^2 + \cdots + F_n^2 \:=\:F_n\cdot F_{n+1}$

Verify $S(1)\!:\;\;F_1^2 \:=\;F_1\cdot F_2$
. . We have: . $1^2 \:=\:1\cdot1$ . . . . . true!

Assume $S(k)$ is true: . $F_1^2 + F_2^2 + F_3^2 + \cdots + F_k^2 \;=\;F_k\cdot F_{k+1}$

We want to show the $S(k+1)$ is true:
. . $F_1^2 + F_2^2 + F_3^2 + \cdots + F_{k+1}^2 \:=\:F_{k+1}\cdot F_{k+2}$

Start with $S(k):$
. . $F_1^2 + F_2^2 + F_3^2 + \cdots + F_k^2 \;=\;F_k\cdot F_{k+1}$

Add $F_{k+1}^2$ to both sides:
. . $F_1^2 + F_2^2+F_3^2 + \cdots + F_k^2 + F_{k+1}^2 \;=\;F_k\cdot F_{k+1} + F_{k+1}^2$

$\text{Factor: }\:F_{k+1}\underbrace{\left(F_k+F_{k+1}\right)}_{\ text{This is }F_{k+2}} \;=\;F_{k+1}\cdot F_{k+2}$

Therefore, we have derived $S(k+1):$
. . $F_1^2 + F_2^2 + F_3^2 + \cdots + F_{k+1}^2 \;=\;F_{k+1}\cdot F_{k+2}$

The inductive proof is complete.

3. Thank you for replying in such short notice.
Now to stare at it till I get it!