$\forall n \in \Bbb N$, I must demonstrate that: $$\sum_{i=0}^n i^4 = \frac15 n^5+\frac12n^4+\frac13n^3-\frac1{30}n $$ $\bullet$ I need to prove that this is true for the first element of the sum (1):

$$\sum_{i=0}^1 i^4 = \frac15 1^5+\frac121^4+\frac131^3-\frac1{30}1 = \frac15 +\frac12+\frac13-\frac1{30} = 1$$

$\bullet$ It is supposed that this is true $\forall n$.

$\bullet$ Now, I need to probe that is true for $n+1$: $$\sum_{i=0}^{n+1} i^4 = \frac15 (n+1)^5+\frac12(n+1)^4+\frac13(n+1)^3-\frac1{30}(n+1) $$

By complete induction method

$$\sum_{i=0}^{n+1} i^4 = \sum_{i=0}^n i^4+(n+1)^4 = \frac15 n^5+\frac12n^4+\frac13n^3-\frac1{30}n + (n+1)^4$$ $$= \frac15 n^5+\frac12n^4+\frac13n^3-\frac1{30}n + (n+1)^2 (n+1)^2$$ $$= \frac15 n^5+\frac12n^4+\frac13n^3-\frac1{30}n + (n^2+2n+1) (n^2+2n+1)$$ $$= \frac15 n^5+\frac12n^4+\frac13n^3-\frac1{30}n + n^4+2n^3+n^2+2n^3+4n^2+2n+n^2+2n+1$$

$$= \frac15 n^5+\frac32n^4+\frac{13}3n^3+6n^2+\frac1{10}n+1$$

But I can not continue to solve this, because I think that is wrong and I do not know what operations must to implemented here.

What have I missed?;

Please put details in the answer if is possible, I do not have a vast knowledge of math.

  • 1
    $\begingroup$ Brian point you are wrong for $\frac{1}{n}$, but the number should be $\frac{119}{30}$, you can find the mistake easily. $\endgroup$ – chenbai Apr 13 '13 at 11:32

Your second bullet point is incorrect: you are not to assume that it’s true for every $n$. Rather, you are to assume as induction hypothesis that it is true for some particular (though unspecified) $n$ and show that it is true for $n+1$. Your calculation is almost right, and you can check it by expanding the desired result and comparing. The desired result is:

$$\begin{align*} \frac15 (n+1)^5&+\frac12(n+1)^4+\frac13(n+1)^3-\frac1{30}(n+1)\\\\ &=\frac15n^5+n^4+2n^3+2n^2+n+\frac15+\\\\ &\qquad+\frac12n^4+2n^3+3n^2+2n+\frac12+\\\\ &\qquad+\frac13n^3+n^2+n+\frac13-\frac1{30}n-\frac1{30}\\\\ &=\frac15n^5+\frac32n^4+\frac{13}3n^3+6n^2+\frac{119}{120}n+1 \end{align*}$$

This differs from your calculation only in the coefficient of $n$, and if you recheck your calculation, you’ll see that you made a mistake there: you should have $-\frac1{30}+2+2=\frac{119}{120}$.


Compute the right hand side of the proposed formula for $n + 1$, and check that it agrees with what you are getting..


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