Question 972088
Assume 1^2 + 2^2 + 3^2 +... n^2 = (n(n+1)(2n+1))/6  is true for all positive integers n. If we replace the right hand side with ((n+2)(n+3)(2n+5))/6  , what term(s) do we add to the left hand side?

I believe the correct answer is (n+1)^2 but I want to double check 
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I'm afraid that's not all but only part of what is added to the left side.

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{{{1^2 + 2^2 + 3^2 +""*""*""*""+n^2}}}{{{""=""}}}{{{(n(n+1)(2n+1))/6}}}

To find out what we have added to the left hand side when we have replaced
the right side with {{{((n+2)(n+3)(2n+5))/6}}} we compare them:

{{{(n(n+1)(2n+1))/6}}}{{{matrix(1,2,compared,to)  }}}{{{((n+2)(n+3)(2n+5))/6}}}  

We see that the first factor on the left numerator is n, while the first
factor on the right numerator is (n+2), so that makes us suspect that n 
has been replaced by (n+2).  

We check to see if that is the case with the second factor on the left. We take
(n+1) and replace n by n+2 in it and we get (n+2+1) or (n+3) which is the second
factor on the right.  So far so good.

We only need to show that if we replace n by n+2 in the third factor on the left
that we will get the third factor on the right (2n+5).  We show that by 
replacing n by n+2 in (2n+1):

(2(n+2)+1) = (2n+4+1) = (2n+5)

Now we know that the right side has been replaced by (n+2) throughout.
Therefore we know that the sum on the left, which is the sum up through n terms,
is now to be carried up to (n+2) terms, which is 2 more terms, so now we
have

{{{1^2 + 2^2 + 3^2 +""*""*""*""+n^2+(n+1)^2+(n+2)^2}}}{{{""=""}}}{{{(n(n+1)(2n+1))/6}}} 

Therefore two more terms {{{(n+1)^2+(n+2)^2}}} have been added to the left,
not just the one term that you were thinking was added.

Edwin</pre>