# Proof by Induction

by Caldus
Tags: induction, proof
 P: 1,004 How did you figure this: $$8*8^k - 3*3^k = 24(8^k - 3^k)$$ ?? You were doing fine up until that point. What you need to do is this: $$8*8^k - 3*3^k = 5*8^k + 3*8^k + 3*3^k = 5*8^k + 3(8^k - 3^k)$$
 P: 1,004 Proof by Induction $$1^2 + 3^2 + 5^2 + ... + (2n - 1)^2 = \frac{n(2n - 1)(2n + 1)}{3} = \frac{4n^3 - n}{3}$$ So for $$k = n + 1$$: $$1^2 + 3^2 + 5^2 + ... + (2n - 1)^2 + (2n + 1)^2 = \frac{(n + 1)(2n + 1)(2n + 3)}{3}$$ $$\underline{1^2 + 3^2 + 5^2 + ... + (2n - 1)^2} + (2n + 1)^2 = \frac{4n^3 + 12n^2 + 11n + 3}{3}$$ We already know it's true for $$n$$ so you can replace the underlined part: $$\frac{4n^3 - n}{3} + (2n + 1)^2 = \frac{4n^3 - n}{3} + 4n^2 + 4n + 1 = \frac{4n^3 + 12n^2 + 11n + 3}{3}$$ Multiply by 3: $$4n^3 - n + 12n^2 + 12n + 3 = 4n^3 + 12n^2 + 11n + 3$$ QED.
 P: 406 I have seen a beautiful geometrical way of deriving the sum of cubes of numbers, i.e. 1^3 + 2^3 + ... + n^3 = (1+2+...n)^2. I wonder if there are simple ways of deriving $$1^2 + 3^2 + 5^2 + ... + (2n - 1)^2 = \frac{n(2n - 1)(2n + 1)}{3}$$
 P: 1,004 You can prove this: $$1^2 + 3^2 + 5^2 + ... + (2n - 1)^2 = \frac{n(2n - 1)(2n + 1)}{3}$$ With a bit of geometry, yes. If you draw squares with a side of 1, 3, 5, etc., one below the other, you can find the sum of their areas with a bit of manipulation, but it's not exactly simple (the idea is simple, the equations are a bit big though).
 P: 1 I have a problem: Knowing that $$1^2 + 2^2 + 3^2 + ... + n^2 = \frac{n(n + 1)(2n + 1)}{6}$$and that $$1^3 + 2^3 + 3^3 + ... + n^3 = \frac{n^2 (n + 1)^2 } {4}$$, calculate $$1^4 + 2^4 + 3^4 + ... + n^4$$. Please, help me!