SOLUTION: 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 th

Algebra ->  Sequences-and-series -> SOLUTION: 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 th      Log On


   

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
Answer by Edwin McCravy(20056) About Me  (Show Source):
You can put this solution on YOUR website!
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%5E2+%2B+2%5E2+%2B+3%5E2+%2B%22%22%2A%22%22%2A%22%22%2A%22%22%2Bn%5E2%22%22=%22%22%28n%28n%2B1%29%282n%2B1%29%29%2F6

To find out what we have added to the left hand side when we have replaced
the right side with %28%28n%2B2%29%28n%2B3%29%282n%2B5%29%29%2F6 we compare them:

%28n%28n%2B1%29%282n%2B1%29%29%2F6matrix%281%2C2%2Ccompared%2Cto%29++%28%28n%2B2%29%28n%2B3%29%282n%2B5%29%29%2F6  

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has been replaced by (n+2).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Now we know that the right side has been replaced by (n+2) throughout.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

%22%22=%22%22%28n%28n%2B1%29%282n%2B1%29%29%2F6 

Therefore two more terms %28n%2B1%29%5E2%2B%28n%2B2%29%5E2 have been added to the left,
not just the one term that you were thinking was added.

Edwin