Help with mathematical assertions for natural numbers

AI Thread Summary
The discussion focuses on proving the assertion that the sum of squares of the first n natural numbers equals n/6 (n+1)(2n+1) using mathematical induction. The user has successfully set up the induction step but is unsure how to proceed with the algebraic manipulation required to complete the proof. To advance, they need to substitute n with (k+1) in the original equation and demonstrate that the resulting expression matches the left side after adding the (k+1)th term. Clarification is provided on how to perform the substitution and simplify the equation. Completing this algebraic step is essential to validate the assertion for all natural numbers n.
MegaDeth
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Homework Statement



Prove by Mathematical Induction that the assertion,
n
∑ r^2 = n/6 (n+1)(2n+1)
r=1

holds for every natural number n.


Homework Equations



Ok, so basically, how do you solve this question? I have got to the Induction step but I'm not sure how to do it.



The Attempt at a Solution



I've replaced n with k so I have,

1^2 + 2^2 + 3^2 + ... k^2 = k/6 (k+1)(2k+1)

Then I've added the (k+1)th term to each side to I have,

1^2 + 2^2 + 3^2 + ... k^2 + (k+1)^2 = k/6 (k+1)(2k+1) + (k+1)^2

So where do I go from here?
 
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Hi MegaDeth! :smile:

Substitute (k+1) in n/6 (n+1)(2n+1) and check if that is equal to your expression.
 
Sorry, but I'm not really sure how to do that :S How do I substitute it and equal it to the expression?
 
To complete the proof, you have to show that the right side of the equation is (k+1)/6 ((k+1)+1)(2(k+1)+1). All it requires is algebra to re-express the right side in a way that shows your assertion is correct.
 
MegaDeth said:
Sorry, but I'm not really sure how to do that :S How do I substitute it and equal it to the expression?

If you substitute n=(k+1) in n/6 (n+1)(2n+1), you get:

(k+1)/6 ((k+1)+1)(2(k+1)+1).


Is that the same as k/6 (k+1)(2k+1) + (k+1)^2?
 
I tried to combine those 2 formulas but it didn't work. I tried using another case where there are 2 red balls and 2 blue balls only so when combining the formula I got ##\frac{(4-1)!}{2!2!}=\frac{3}{2}## which does not make sense. Is there any formula to calculate cyclic permutation of identical objects or I have to do it by listing all the possibilities? Thanks
Since ##px^9+q## is the factor, then ##x^9=\frac{-q}{p}## will be one of the roots. Let ##f(x)=27x^{18}+bx^9+70##, then: $$27\left(\frac{-q}{p}\right)^2+b\left(\frac{-q}{p}\right)+70=0$$ $$b=27 \frac{q}{p}+70 \frac{p}{q}$$ $$b=\frac{27q^2+70p^2}{pq}$$ From this expression, it looks like there is no greatest value of ##b## because increasing the value of ##p## and ##q## will also increase the value of ##b##. How to find the greatest value of ##b##? Thanks

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