SOLUTION: PLEASE HELP !!!!! In using mathematical induction to prove 1^2+3^2+5^2=...+(2n-1)^2 = (n(2n-1)(2n+1)/3), the P k statment is 1^2+3^2+5^2+...+(2k-1)^2 = (k(2k-1)(2k+1)/3) Is thi

Question 144253: PLEASE HELP !!!!!
In using mathematical induction to prove 1^2+3^2+5^2=...+(2n-1)^2 =
(n(2n-1)(2n+1)/3), the P k statment is 1^2+3^2+5^2+...+(2k-1)^2 = (k(2k-1)(2k+1)/3)
Is this true or false
THANK YOU
Answer by stanbon(75887) (Show Source): You can put this solution on YOUR website!
In using mathematical induction to prove
1^2+3^2+5^2=...+(2n-1)^2 = (n(2n-1)(2n+1)/3)
----------------------
the P(k) statment is 1^2+3^2+5^2+...+(2k-1)^2 = (k(2k-1)(2k+1)/3)
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Comment:
Proof by math induction requires
1st: that you show it is true for some number, n.
2nd: you then assume it is true for n=k
3rd: Based on your assumption you must show it is true for n=k+1.
------------------------
Your Problem:
Is it true for n=1 ?
1^2 = 1*(2-1)*(2+1)/3 = 1*1*3/3 = 1
Ans: Yes it is true for n=1
---------------------
Assume it is true for n=k:
P(k) = 1^2+3^2+5^2+...+(2k-1)^2 = (k(2k-1)(2k+1)/3)
----------------------
Show it is true for n=k+1:
P(k+1) = 1^2+3^3+5^2+...+(2k-1)^2 + (2[(k+1)-1])^2 = [(k+1)(2(k+1-1)(2(k+1)+1)]/3
P(K+1) = 1^2+3^3+5^2+...+(2k-1)^2 + (2[(k+1)-1])^2 = [(k+1)(2K)(2k+3)]/3
----------
Substitute into the left side of P(k+1) to get:
P(k) + (2(k+1)-1])^2
But P(k) = (k(2k-1)(2k+1)/3); so you get
= (k(2k-1)(2k+1)/3)+ (2(k+1)-1])^2
= (k(2k-1)(2k+1)/3)+ (2k+1)^2
= [(k(4k^2-1)] + 3*(4k^2+4k+1)/3
= [K(4K^2-1) + 12K^2+12k+3]/3
= [4k^3+12k^2+11k+3]/3
==========
But the right side of P(k+1) must be
[(k+1)(2K)(2k+3)]/3
= (2k)(2k^2+5k+3)
= [4k^3 + 10k^2 + 6k]/3
Conclusion: Form is false
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Comment: I can only hope I have not made a
mistake in the algebric manipulation.
The process is correct; the algebra could
use checking.
--------------
Cheers,
Stan H.
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