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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?

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  • $\begingroup$ The $n$ must be a positive integer. Check it again $\endgroup$ – Isaac YIU Math Studio Oct 18 '19 at 16:07
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    $\begingroup$ 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. $\endgroup$ – J.G. Oct 18 '19 at 16:16
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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.

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    $\begingroup$ In 1+4*2*√x, what is it equal to, or is there an inequality present. What is the other condition? $\endgroup$ – Aditya Oct 18 '19 at 16:25
  • $\begingroup$ 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. $\endgroup$ – BJKShah Oct 18 '19 at 16:38
  • $\begingroup$ Yes of course. Thanks! $\endgroup$ – Aditya Oct 18 '19 at 16:56
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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.

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$\sum_{k=1}^{n}k^3$ 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.

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