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Here are two famous general formulas: 1 + 2 + 3 + ... + n = n(n+1)/2 and 12 + 22 + 32 +..+N2 = n(n + 1) (2n + 1)/6. For example. 1 + 2 + 3 = 6= 4-3 / 2. Works! What about l3 + 23 + 33 + ... + n3 =? Let's see whether we can make an educated guess. Expanding the right sides of our first two formulas gives So, the formula for the first sum is of the form an2 + bn + c and the formula for the second sum is of the form an3 + bn2 + cn + d (with just the right choices of coefficients). So, maybe, the next formula is of the form 13 + 23 + 33 + ... + n3 = an4 + bn3 + cn2 + dn + e. Here then is your mission: By assuming that this can really be done, find coefficients a, b, c, d, e that work. To answer this question you may want to use the Linear Algebra Toolkit (available online - google it) or other software. Just make sure that you record the input and output and justify what you are doing. (Hint: Substitute a few different values for n in this would-be formula to generate some linear equations in the unknowns a.b,c,d,e... Also, to make it a bit easier for yourself, note that the first two formulas even work for n = 0, if you interpret the sums on the left to add up to 0. So this should also work here.) (Bonus: 10 points) Prove that your formula really works. (Hint: One standard way of proving formulas like this is called Proof by Induction, You can learn how to do such a proof here http://nrich.maths.org/4718, but we'll also give a mini-introduction to this in class).