SOLUTION: use mathematical induction to prove that
1^2 + 2^2 + 3^2 +...+ n^2 = n(n+1)(2n+1)/6
for all positive integral values of n
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Question 229820: use mathematical induction to prove that
1^2 + 2^2 + 3^2 +...+ n^2 = n(n+1)(2n+1)/6
for all positive integral values of n
Answer by solver91311(24713) (Show Source): You can put this solution on YOUR website!
prove
(2n+1)}{6})
1. Show that
is true for
and
(2(1)+1)}{6}\ =\ \frac{6}{6}\ = 1)
2. Assume
is true for
some positive integer 
, then show
the relationship is true for 
, namely that:
((n+1)+1)(2(n+1)+1)}{6})
First note that:
^2)
which can be written:
(2n+1)}{6}\ +\ (n + 1)^2)
because we assumed the relationship to be true for some positive integer .
But
(2n+1)}{6}\ +\ (n + 1)^2\ =\ \frac{2n^3+9n^2+13x+6}{6})
(Verification of the previous step is left as an exercise for the student)
And
((n+1)+1)(2(n+1)+1)}{6}\ =\ \frac{2n^3+9n^2+13x+6}{6})
(Verification of the previous step is also left as an exercise for the student)
Therefore, if the statement is true for some , it must be true for . Since it was proven true for 
, it must be true for 
, then it must be true for 
...
John
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