Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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!)

share|improve this question
1  
en.wikipedia.org/wiki/Binomial_expansion has all you need. –  msh210 Nov 23 '11 at 8:15
2  
$$(x+y)^n=\sum\limits_{k=0}^n \binom{n}{k} x^k y^{n-k}$$ is what you need. –  J. M. Nov 23 '11 at 8:16

2 Answers 2

up vote 2 down vote accepted

$(2a-1)^n$ 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 $n$.

share|improve this answer
    
Could you also please show me how to solve $-n 2^{n-1}=192$? Thanks! –  tina nyaa Nov 23 '11 at 9:02
    
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 $n=6$. –  David Mitra Nov 23 '11 at 9:06
    
Re "best to just guess": cf. my comment on my answer. –  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 $n=6$.

share|improve this answer
    
Thanks! Just on two little parts, how would you solve $-n\cdot2^{n-1}=-192$ and how did you get $\frac{n!}{(n-1)!}(2a)^{n-1}=-n\cdot2^{n-1}a^{n-1}$? –  tina nyaa Nov 23 '11 at 8:57
    
@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 $(2a)^m=2^ma^m$. –  msh210 Nov 23 '11 at 15:35

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.