I disgraced myself in the eyes of my ancestors
Just now, a young friend of mine asked of me some wisdom for O level maths. I thought, how hard could that be? So I agreed to take a look at it. Binomial theorem question. Cool, I hadn't touched it in, oh, more than ten years, but hey, I'm God, right? I can do it all. Convinced of my own awesome ability to assimilate and apply, I wiki-ed binomial theorem, and here's what I found.



Great, I thought, given the formula, was there any related problem I could not solve? Fuck, no! So here's the question (more or less). Given (2+x)^6, find the first three coefficients of its binomial expansion in ascending degrees of x and, hence or otherwise, find the values of p and q in the equation:

(2 + px -x^2)^6 = 64 + qx + 48x^2 + ...

The first thing I did was to rearrange the original expression like so.

(x+2)^6

This would, when I applied the binomial theorem, expand nicely into ascending degrees of x, right? So, I expanded the first 3 terms, and I got

64 + 192x + 240x^2

Cool. Easy part over. So far so good. Then the shit hit the fan. I tried to factorise 2+px-x^2 into something resembling either x+2 or 2+x. I will not go into the gory details of the exotic methods I used or the wrong turns I made. Suffice it to say that trying to do math problems with only your memory, a text edit buffer and the calculator in your computer while IM-ing other people simultaneously is never a good idea.

You seldom see the whole picture.

By then, my friend was getting visibly discontented with my lack of omniscience and I bade the sprout to go to sleep, feeling that I had disgraced myself in the eyes of my ancestors with my lack of brilliance before a minor.

After that, I finally got out some paper and wrote the question down, determined to subdue the dragons of ignorance once and for all. I noticed something immediately.

(2 + px - x^2)^6 = (px - x^2 + 2)^6 = ((px - x^2) + 2)^6

Expand LHS using binomial theorem treating (px - x^2) as one variable, you get:

(px - x^2 + 2)^6 = 64 + 192(px - x^2) + 240(px - x^2)^2 + . . .
(px - x^2 + 2)^6 = 64 + 192px - 192x^2 + 240px^2 + (unimportant stuff where the power of x is greater than 2)

Equating LHS and RHS, you get the following:

64 + 192px - 192x^2 + 240px^2 + . . . = 64 + qx + 48x^2 + . . .

Comparing coefficients of x^2, you have

240p - 192 = 48 => p = 1

Substituting p = 1 and comparing coefficients of x, we get

192 = q

Forgive me, O honourable ancestors, for I have disgraced myself by failing to notice the obvious immediately. As penance, I shall stop declaring that I am God for half a day and promise to always write the fucking problem on a fucking piece of paper with a fucking pen.

p.s. If you're reading this, kid, trust me. Write it down on a piece of paper. It looks way clearer.  ¶
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