# Induction to prove

• Aug 22nd 2009, 06:40 PM
lin.13579
Induction to prove
Hi, could someone please explain how to do this:

use induction to prove that
$\displaystyle \sum_{i=1}^n i^2 = \frac{n(n+1)(2n+1)}{6}$

for all positive integer n.

[Note: $\displaystyle \sum_{i=1}^n i^2$ means the sum $\displaystyle 1^2+2^2+3^2+...+n^2$]

thank you!!!
• Aug 22nd 2009, 07:02 PM
mr fantastic
Quote:

Originally Posted by lin.13579
Hi, could someone please explain how to do this:

use induction to prove that
$\displaystyle \sum_{i=1}^n i^2 = \frac{n(n+1)(2n+1)}{6}$

for all positive integer n.

[Note: $\displaystyle \sum_{i=1}^n i^2$ means the sum $\displaystyle 1^2+2^2+3^2+...+n^2$]

thank you!!!

Amazing what you can find in a few seconds using Google. Key words: proof induction sum squares.

Read this (one of 747,000 found in 0.28 seconds): http://math.arizona.edu/~kerl/doc/induction.pdf
• Aug 22nd 2009, 07:06 PM
lin.13579
Quote:

Originally Posted by mr fantastic
Amazing what you can find in a few seconds using Google. Key words: proof induction sum squares.

Read this (one of 747,000 found in 0.28 seconds): http://math.arizona.edu/~kerl/doc/induction.pdf

oh, my gosh , thank you , I wrote this eqution for a long time......(LaTex). thanks
• Aug 22nd 2009, 07:55 PM
quah13579
Mathematical Induction Example 2 --- Sum of Squares

Problem: For any natural number n , 1^2 + 2^2 + ... + n^2 = n( n + 1 )( 2n + 1 )/6.

Proof:
Basis Step: If n = 0, then LHS = 02 = 0, and RHS = 0 * (0 + 1)(2*0 + 1)/6 = 0 .
Hence LHS = RHS.
Induction: Assume that for an arbitrary natural number n,
1^2 + 2^2 + ... + n^2 = n( n + 1 )( 2n + 1 )/6. -------- Induction Hypothesis
To prove this for n+1, first try to express LHS for n+1 in terms of LHS for n, and use the induction hypothesis.
Here let us try
LHS for n + 1 = 1^2 + 2^2 + ... + n^2 + (n + 1)^2 = ( 1^2 + 2^2 + ... + n^2 ) + (n + 1)^2
Using the induction hypothesis, the last expression can be rewritten as
n( n + 1 )( 2n + 1 )/6 + (n + 1)^2
Factoring (n + 1)/6 out, we get
( n + 1 )( n( 2n + 1 ) + 6 ( n + 1 ) )/6
= ( n + 1 )( 2n^2 + 7n + 6 )/6
= ( n + 1 )( n + 2 )( 2n + 3 )/6 ,
which is equal to the RHS for n+1.

Thus LHS = RHS for n+1.

End of Proof.
• Aug 26th 2009, 11:51 PM
yoonsi
Apologies for asking so many questions, but I'm trying to master as many questions as I can, but why does "( 2n + 3 )" represent the next case for (2n+1)? shouldnt it be (2n+2)?
• Aug 27th 2009, 04:19 AM
Defunkt
The next case for $\displaystyle n$ is $\displaystyle n+1$, so the next case for $\displaystyle 2n + 1$ is $\displaystyle 2(n+1) + 1 = 2n + 3$