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)^
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-> 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)^
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Question 1080517: Using mathematical induction, show its true for all natural #'s n + =
So far I got
Proof n=1 =
1 = 3/3
1 = 1
so its true for n=1
next assume n = k, so
+ =
then we must prove n = k+1
+ + =
we do some replacing with our n = k
+ =
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,
+ = + =
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(52778) (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
+ =
So far I got
Proof n=1
=
1 = 3/3
1 = 1
so its true for n=1
next assume n = k, so
+ =
then we must prove n = k+1
+ + = <<<---Corrected
we do some replacing with our n = k
+ = <<<---Corrected
we have our left hand side that we must match to our right hand side
+ = <<<---Corrected
+ = <<<---Corrected
Below I continue the proof.
Now let us transform the left side step by step:
+ =
+ =
=
= . (*)
Now notice that = , and therefore you can continue the chain of equalities (*) in this way
= = ,
and this is EXACTLY our RIGHT side.
