SOLUTION: Using mathematical induction, show its true for all natural #'s n {{{ 1^2+3^2+5^2+ ...}}}+ {{{ (2n-1)^2 }}} = {{{ (n(2n-1)(2n+1))/3 }}} So far I got Proof n=1 {{{ (2(1)-1)^

Algebra ->  Proofs -> SOLUTION: Using mathematical induction, show its true for all natural #'s n {{{ 1^2+3^2+5^2+ ...}}}+ {{{ (2n-1)^2 }}} = {{{ (n(2n-1)(2n+1))/3 }}} So far I got Proof n=1 {{{ (2(1)-1)^      Log On


   

Question 1080517: Using mathematical induction, show its true for all natural #'s n
+1%5E2%2B3%5E2%2B5%5E2%2B+...+ +%282n-1%29%5E2+ = +%28n%282n-1%29%282n%2B1%29%29%2F3+
So far I got
Proof n=1
+%282%281%29-1%29%5E2+ = +%281%282%281%29-1%29%282%281%29%2B1%29%29%2F3+
1 = 3/3
1 = 1
so its true for n=1
next assume n = k, so
+1%5E2%2B3%5E2%2B5%5E2%2B+...+ +%282k-1%29%5E2+ = +%28k%282k-1%29%282k%2B1%29%29%2F3+
then we must prove n = k+1
+1%5E2%2B3%5E2%2B5%5E2%2B+...+ +%282k-1%29%5E2+ + +%282k%2B1-1%29%5E2+ = +%28k%2B1%282k%2B1-1%29%282k%2B1%2B1%29%29%2F3+
we do some replacing with our n = k

+%28k%282k-1%29%282k%2B1%29%29%2F3+ + +%282k%2B1-1%29%5E2+ = +%28k%2B1%282k%2B1-1%29%282k%2B1%2B1%29%29%2F3+
we have our left hand side that we must match to our right hand side, but this is where my math get's messy when I go to simplify,
+%28k%282k-1%29%282k%2B1%29%29%2F3+ + +%282k%2B1-1%29%5E2+ = +%282k%28k%2B1%29%282k%2B2%29%29%2F3+
+%28k%282k-1%29%282k%2B1%29%29%2F3+ + +4k%5E2+ = +%282k%28k%2B1%29%282k%2B2%29%29%2F3+
now I'm left with a +4k^2 and I'm unsure what method to use here, or maybe I did the first steps wrong? If I factor the k(2k-1)(2k+1) I get a 4k^3-k and I thought maybe multiply by (3/3) to get everything as denominator of 3, but it doesn't really help me understand what I should do next for certain. Any help would be greatly appreciated
Answer by ikleyn(53937) About Me  (Show Source):
You can put this solution on YOUR website!
.
I start reproducing your post; then make some correction of your writing and then complete the proof. 

Using mathematical induction, show its true for all natural #'s n 
+1%5E2%2B3%5E2%2B5%5E2%2B+...+ +%282n-1%29%5E2+ = +%28n%282n-1%29%282n%2B1%29%29%2F3+

So far I got 
Proof n=1
+%282%281%29-1%29%5E2+ = +%281%282%281%29-1%29%282%281%29%2B1%29%29%2F3+
1 = 3/3
1 = 1
so its true for n=1

next assume n = k, so 

+1%5E2%2B3%5E2%2B5%5E2%2B+...+ +%282k-1%29%5E2+ = +%28k%282k-1%29%282k%2B1%29%29%2F3+

then we must prove n = k+1

+1%5E2%2B3%5E2%2B5%5E2%2B+...+ +%282k-1%29%5E2+ + +%282%28k%2B1%29-1%29%5E2+ = +%28%28k%2B1%29%2A%282%28k%2B1%29-1%29%282%28k%2B1%29%2B1%29%29%2F3+        <<<---Corrected

we do some replacing with our n = k 
 
 +%28k%282k-1%29%282k%2B1%29%29%2F3+ + +%282%28k%2B1%29-1%29%5E2+ = +%28%28k%2B1%29%2A%282k%2B1%29%282k%2B3%29%29%2F3+                       <<<---Corrected

we have our left hand side that we must match to our right hand side             

 +%28k%282k-1%29%282k%2B1%29%29%2F3+ + +%282%28k%2B1%29-1%29%5E2+ = +%28%28k%2B1%29%2A%282k%2B1%29%282k%2B3%29%29%2F3+                       <<<---Corrected

 +%28k%282k-1%29%282k%2B1%29%29%2F3+ + +%282k%2B1%29%5E2+ = +%28%28k%2B1%29%2A%282k%2B1%29%282k%2B3%29%29%2F3+                           <<<---Corrected

Below I continue the proof. 

Now let us transform the left side step by step:

+%28k%282k-1%29%282k%2B1%29%29%2F3+ + +%282k%2B1%29%5E2+ = 

+%28k%282k-1%29%282k%2B1%29%29%2F3  + %283%2A%282k%2B1%29%5E2%29%2F3+ = 

+%28%282k%2B1%29%2A%28k%2A%282k-1%29+%2B+3%2A%282k%2B1%29%29%29%2F3 = 

%28%282k%2B1%29%2A%282k%5E2-k+%2B+6k+%2B+3%29%29%2F3 = %28%282k%2B1%29%2A%282k%5E2+%2B+5k+%2B+3%29%29%2F3.   (*)

Now notice that 2k%5E2+%2B+5k+%2B+3 = %28k%2B1%29%2A%282k%2B3%29,  and therefore you can continue the chain of equalities (*) in this way

= %28%282k%2B1%29%2A%28k%2B1%29%2A%282k%2B3%29%29%2F3 = %28%28k%2B1%29%2A%282k%2B1%29%2A%282k%2B3%29%29%2F3,

and this is EXACTLY our RIGHT side.

The proof is completed.


-----------
There is a free of charge textbook in this site
    ALGEBRA-II - YOUR ONLINE TEXTBOOK.

There is a chapter in this textbook Method of Mathematical induction.

There are the lessons in this chapter
    - Mathematical induction and arithmetic progressions
    - Mathematical induction and geometric progressions
    - Mathematical induction for sequences other than arithmetic or geometric (**)
    - Proving inequalities by the method of Mathematical Induction

that might be interested to you, especially the lesson marked (**).


Happy learning !!