Hi, could someone please explain how to do this:

use induction to prove that

for all positive integer n.

[Note: means the sum ]

thank you!!!

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

use induction to prove that

for all positive integer n.

[Note: means the sum ]

thank you!!! - Aug 22nd 2009, 07:02 PMmr 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 - Aug 22nd 2009, 07:06 PMlin.13579
- Aug 22nd 2009, 07:55 PMquah13579
**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, then*n*=*0*, and*LHS*=*02*=*0**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 forfirst try to express*n*+*1*,for*LHS*in terms of*n*+*1*for*LHS*and use the induction hypothesis.*n*,

Here let us try

for*LHS**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**(**out, we get*n + 1*)/6

**(***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 thefor*RHS**n*+*1*.

Thusfor*LHS*=*RHS**n*+*1*.

**End of Proof.** - Aug 26th 2009, 11:51 PMyoonsi
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 AMDefunkt
The next case for is , so the next case for is