How do I prove that for any natural number $n$ we have $$\sum_{i=0}^n i^4 \neq \left(\sum_{i=0}^n i\right)^3?$$
Any help would be greatly appreciated.
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Recall that $$\sum_{i=1}^n i^4 = \dfrac{n(n+1)(2n+1)(3n^2+3n-1)}{30}$$ and $$\sum_{i=1}^n i = \dfrac{n(n+1)}2$$ We hence need $$\left(\dfrac{n(n+1)}2\right)^3 = \dfrac{n(n+1)(2n+1)(3n^2+3n-1)}{30}$$ This gives us either $n(n+1)=0$ or $$15(n(n+1))^2 = 4(2n+1)(3n^2+3n-1) \,\,\,\, (\spadesuit)$$ $(\spadesuit)$ can be simplified as $$15n^4+6n^3-21n^2-4n+4 = 0 \implies (n-1)(15n^3+21n^2-4) = 0 \,\,\,\, (\clubsuit)$$ Now consider the function $f(n) = (15n^3+21n^2-4)$. Note that we have $f'(n) = 45n^2+42n >0 $ for all $n \geq 1$. Hence, the function $f(n)$ is increasing for all positive integers. Further, $f(1) = 32 > 0$. Hence, $f(n)$ has no non-negative integer as its roots. The only solution to your initial problem is $n=0$ and $n=1$.
Adhvaitha
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Your statement is false. If you use $n=1$, both sides of your statement equal $1$. For that matter, if $n=0$ then both sides are zero, but you may not consider zero to be a natural number.
It is true that the statement is true for $n>1$. Is that what you want to prove? If so, find the polynomial expressions for each side. Subtracting one side from the other leads you to search for the roots of a sixth-degree polynomial. Show that two of the roots are $0$ and $1$, three roots are negative, and one root is between zero and one. That accounts for all six roots, so there is no root for $n>1$. Or, as @Deepak suggests, use the rational root theorem to find all rational roots, which are just $-1,0,1$.
Rory Daulton
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Here's the question: You are asked to prove whether for any natural number n: sum for i = 0 to n of i^4 = (sum for i = 0 to n of i)^3
Use mathematical induction.
– Ian Moon May 01 '15 at 13:06 -
@IanMoon: If that is your question, use $n=1$ and perhaps $n=0$ to show that equality does hold "for any natural number $n$". You probably can use induction to show that the left hand side is greater than the right hand side for $n>1$. – Rory Daulton May 01 '15 at 13:09
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But if I use n = 0 (or n=1), both LHS and RHS will result in 0 (or 1), meaning that the equality holds. I need to disprove that they aren't equal to each other. – Ian Moon May 01 '15 at 13:15
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@IanMoon: I don't get it. Using $0$ or $1$ proves equality, which does "disprove that they aren't equal to each other." – Rory Daulton May 01 '15 at 13:16
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Oh sorry, it's the other way around. I have to prove that they aren't equal to each other. – Ian Moon May 01 '15 at 13:21
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@IanMoon: Your latest edit of your question removed the "by induction" phrase. Do you have to use mathematical induction or not? Also, my one comment was wrong: the left hand side is less than the right hand side for $n>1$. Have you tried proving that by induction? – Rory Daulton May 01 '15 at 13:30
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Yes, I have to use the mathematical induction. What do you mean by $n>1$? Do I have to apply, for example, n =2 in the LHS and see if it matches the RHS? – Ian Moon May 01 '15 at 13:44
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As Rory Daulton has remarked we have $L(0)=R(0)$ and $L(1)=R(1)$. But then $L(2)=17< 27=R(3)$. Therefore it is sufficient to prove $$L(n)-L(n-1)\leq R(n)-R(n-1)\qquad(n\geq3)\ ,$$ which is the same as $$n^4\leq \left({n(n+1)\over2}\right)^3-\left({(n-1)n\over2}\right)^3=n^3\ {6n^2+2\over 8}\qquad(n\geq3)\ .$$ This amounts to $4n\leq3n^2+1$, resp., to $$(3n-1)(n-1)\geq0\qquad(n\geq3)\ ,$$ which is obviously true.
Christian Blatter
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Don't know where the attachment went off.
– Ian Moon May 01 '15 at 12:41