In the expansion of (2a − 1)^n, the coefficient of the second term is −192. Find the value of n.

How would I work this question out, without brute-forcing every combination?

(I got a shocking high school education, which is something I'm only starting to come to terms with now that I'm no longer inside its walls, so please explain your answer nicely!)


$(2a-1)^n$ has the form $$ \underbrace{(2a-1)(2a-1)(2a-1)\cdots(2a-1)}_{n-\rm terms} $$

To obtain the product, you would keep applying the distributive law, again and again, until there were no multiplications to perform. For instance $$\color{darkgreen}{(2a-1)}\color{darkblue}{(2a-1)}=2a\color{darkblue}{(2a-1)}-1 \color{darkblue}{(2a-1)}=\color{maroon}{2a\cdot2a-2a-2a+1}=4a^2-4a+1.$$

Please note that the maroon expression above could have been obtained by summing all products obtained from selecting a term (either $2a$ or $-1$) of the green expression on the left and a term of the blue expression on the left to form the product (this is why we have the so-called "FOIL" method; the mnemonic gives all possible products, here).

Going back to $$ \tag{1}(2a-1)(2a-1)(2a-1)\cdots(2a-1), $$ all these multiplications arising from all that distributing would give in the end (after collecting like terms) an expression of the form: $$ a_na^n+a_{n-1}a^{n-1}+\cdots +a_1a+a_0 $$ where the $a_i$ are particular numbers.

As in the case above, one can find $(2a-1)^n$, by taking the sum of all possible products obtained by selecting either $2a$ or $-1$ from each of the $n$ factors of $(1)$ to form the product.

Now, how would one get the $a_{n-1}a^{n-1}$ term?

Well, to obtain "part" of it you'd pick a "$\color{darkgreen}{-1}$" from one of the terms in $(1)$ and $\color{maroon}{2a}$ from the others. For example: pick $$ (\color{maroon}{2a}{-1})(\color{maroon}{2a}-1)({2a}\color{darkgreen}{-1})\cdots(\color{maroon}{2a}-1), $$ and then take the product: $(2a)^{n-1}( -1) $.

But, there are $n $ ways of selecting the $-1$ and each of these gives a $(2a)^{n-1}(-1) $ term. Adding all these together gives the $a_{n-1}a^{n-1}$ term, so: $$ a_{n-1}a^{n-1}=n(2a)^{n-1}(-1) =-n2^{n-1}a^{n-1}. $$

So $a_{n-1}=-n 2^{n-1}$.

Finally set $-n 2^{n-1}=-192$ and solve for $n$.

  • $\begingroup$ Could you also please show me how to solve $-n 2^{n-1}=192$? Thanks! $\endgroup$ – tina nyaa Nov 23 '11 at 9:02
  • $\begingroup$ It's best to just guess and check here... Maybe it's $n=5$? Well $-5\cdot 2^{5-1}=-80$, so no.. $n=6$? Well, $-6\cdot 2^{6-1}=-192$, so it's $n=6$. $\endgroup$ – David Mitra Nov 23 '11 at 9:06
  • $\begingroup$ Re "best to just guess": cf. my comment on my answer. $\endgroup$ – msh210 Nov 23 '11 at 19:43

See http://en.wikipedia.org/wiki/Binomial_expansion for an explanation; a summary is that the $(k+1)$th term of the expansion is $\frac{n!}{(n-k)!k!}(2a)^{n-k}(-1)^k$. Thus the second term is $\frac{n!}{(n-1)!1!}(2a)^{n-1}(-1)^1=-n\cdot2^{n-1}a^{n-1}$, with coefficient $-n\cdot2^{n-1}$. Since we know that's $-192$, we solve $-n\cdot2^{n-1}=-192$ for $n$ and get $n=6$.

  • $\begingroup$ Thanks! Just on two little parts, how would you solve $-n\cdot2^{n-1}=-192$ and how did you get $\frac{n!}{(n-1)!}(2a)^{n-1}=-n\cdot2^{n-1}a^{n-1}$? $\endgroup$ – tina nyaa Nov 23 '11 at 8:57
  • $\begingroup$ @tinanyaa: Re solving $-n\cdot2^{n-1}=-192$: Since you're (I assume) given $n$ is an integer, you can factor $192$, see $3$ goes into it, guess $n=6$, and test whether it works. Re the other expression: Note $\frac{n!}{(n-1)!}=n$ and $(2a)^m=2^ma^m$. $\endgroup$ – msh210 Nov 23 '11 at 15:35

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