# The value which is not possible for $\sum_{k=1}^n k^3$ is

a)25

b)36

c)225

d)441

The value becomes $$\left[\frac{(n)(n+1)}{2}\right]^2$$ Since all options are perfect squares, I don’t really know what to do with This.

The discrimaint ie. $$b^2-4ac>0$$ for all options if we solve the quadratic equation for each option.

The right answer is $$25$$. How do we get that?

• The $n$ must be a positive integer. Check it again – Isaac YIU Math Studio Oct 18 '19 at 16:07
• What have you tried? You could literally have just added the first three cubes to realise the answer is a), since the partial sums will just keep growing. – J.G. Oct 18 '19 at 16:16

Let $$\left[\frac{(n)(n+1)}{2}\right]^2=x$$ It is quadratic in $$n$$. $$n^2+n-2\sqrt{x}=0$$ The discriminant $$1+4*1*2\sqrt{x}=1+8\sqrt{x}$$ should be a perfect square because $$n$$ must be whole number. So, number need not only be a square number. It should satisfy the above condition as well. $$25$$ doesn't satisfy it and hence the answer.
• Discriminant $b^2-4ac$ as you mentioned must be a either $0$ or perfect square if the quadratic equation is to have an integer as a solution. – BJKShah Oct 18 '19 at 16:38
The expression inside the square is a triangular number. Thus we take the square roots of the choices and see which is not a triangular number, and only the square root of (a) is not so: $$5$$ as compared to $$6,15,21$$.
Indeed, $$36,225,441$$ are the sums of the first $$3,5,6$$ cubes respectively.
$$\sum_{k=1}^{n}k^3$$ is always the square of a triangular number, and $$m$$ is a triangular number iff $$8m+1$$ is a square. It turns out that options b,c,d are fine (they correspond to $$n=3,5,6$$), but a is not since $$8\cdot 5+1$$ is not a square.