enter image description here

How do i start this? Any hints will be appreciated


  • 2
    $\begingroup$ Hint: the sum of the coefficients is the value at $x=1$ which is $3^n$. $\endgroup$
    – dxiv
    Sep 25, 2016 at 19:10
  • $\begingroup$ Can you elaborate? $\endgroup$
    – J. Deff
    Sep 25, 2016 at 19:16
  • 3
    $\begingroup$ Ok I'm confused by this problem. If you expand this out for a positive integer power, you get each degree of $x$ from $-1$ to $-2n$, plus you have a constant term. So there will always be $2n+1$ terms in the expansion. I'm not seeing how you can ever get $28$ terms. I even checked with Wolfram for $n=13$ and $14$ and they have $27$ and $29$ terms respectively. $\endgroup$
    – pancini
    Sep 25, 2016 at 19:19
  • 2
    $\begingroup$ @Elliot: And if they mean the number of terms before like terms are collected, the number of terms must be a power of $3$. (Moreover, the available answers imply that $n$ is $5,6$, or $7$, none of which is consistent with either reading.) $\endgroup$ Sep 25, 2016 at 19:26
  • $\begingroup$ After expansion you'll get a sum of terms in the form $\frac{a_k}{x^k}$ where $a_k$ is the coefficient. For $x=1$ the sum gives the sum of the coefficients. $\endgroup$
    – dxiv
    Sep 25, 2016 at 19:28

3 Answers 3


Few months ago, a student brought this question to my attention. It baffled me for a while (The number of terms is given as 28. Is it before/after gathering terms with powers of x? If you gather, you only get 13 terms).

It appeared in IIT-2016-Mains entrance examination (Indian Institute of Technology). A student gets 1 or 2 minutes to answer the question.

A student is expected to use Trinomial Expansion, without gathering terms (That makes it ill-formed).

In Trinomial Expansion, the number of terms is given out as:

Thus n=6.
Th sum of coefficients is given as: $3^n=729$

Here is the website, question paper and answer key

Explanation on how gathering reduces number of terms: There are 28 terms in the full trinomial expansion. A lot of terms can be combined, with similar powers of x.

For ex., consider the term with $\displaystyle \frac{1}{x^4}$ in it. This can be obtained from the following three terms:

$\displaystyle \binom{6}{4,0,2} \,\, (1)^4 \,\, \left(\frac{-2} {x}\right)^0 \,\, \left(\frac{4}{x^2}\right)^2 \,\, +$
$\displaystyle \binom{6}{3,2,1} \,\, (1)^3 \,\, \left(\frac{-2}{x}\right)^2 \,\, \left(\frac{4}{x^2}\right)^1 \,\, +$
$\displaystyle \binom{6}{2,4,0} \,\, (1)^2 \,\, \left(\frac{-2}{x}\right)^4 \,\, \left(\frac{4}{x^2}\right)^0$

$\displaystyle = \frac{1440}{x^4}$

The above uses the notation of trinomial coefficients as follows:

$\displaystyle \binom{n}{k_1,k_2,k_3}$ is equal to $\displaystyle \frac{n!}{k_1!\,\,k_2!\,\,k_3!}$

When you simplify all the terms, you get only 13 terms (by summarizing 28 terms):

\begin{equation} \\ 1 - \frac{12}{x} + \frac{84}{x^2} - \frac{400}{x^3} + \frac{1440}{x^4} - \frac{4032}{x^5} + \frac{9024}{x^6} - \frac{16128}{x^7} + \frac{23040}{x^8} - \frac{25600}{x^9} + \frac{21504}{x^{10}} - \frac{12288}{x^{11}} + \frac{4096}{x^{12}} \end{equation}

  • 1
    $\begingroup$ Good answer as it points to a most plausible mistake in the stated problem. It would be nice if you could strikingly restate the problem with the correct wording (by adding trinomial) and everything has been settled. (+1) $\endgroup$ Oct 19, 2016 at 19:40
  • $\begingroup$ Will this formula fails when some terms will cancel out? I have seen trinomial expansions with terms $2n+1$ $\endgroup$
    – J. Deff
    Oct 20, 2016 at 6:29
  • $\begingroup$ @J.Deff, I don't have terms that cancel out exactly (i.e., like +2-2=0). But, terms do get summarized (i.e., like +4-2=2). That gives us 13 terms, which results in fractional value for 'n'. Thus the question makes sense only when we interpret it as The number of terms in the trinomial expansion of .... is 28. Find the sum of the coefficients in the expansion $\endgroup$
    – blackpen
    Oct 20, 2016 at 20:05

I fully agree with all concerns expressed in the previous answer and comments. Surely this question is formulated in an unclear manner. In my opinion, a possibility is that who conceived this problem did not refer to the actual coefficients resulting by assigning to $x $ specific values, but simply to the generic binomial coefficients resulting from the expansion of an expression of the form $(a+b+c)^n $ . If we interpret the question in this way, we get that, for any given $n $, there are:

  • $3$ terms containing only one variable (i.e., $a^n $, $b^n $, and $c^n $);

  • $3 (n-1)$ terms containing two variables. This is obtained by giving, for each of the $3$ possible pairs $ab $, $ac $, $bc $, all possible exponents $i,n-i$ (with $1 \leq i \leq n-1$) to the two variables;

  • $1/2 \, (n-1)(n-2)$ terms containing all three variables. This is obtained by giving, for any given exponent $j$ of the first variable (with $1 \leq j \leq n-2$), all possible $n-j-1$ exponents to the other two variables, thus leading to the sum $\sum_{j=1}^{n-2} n-j-1$, which in turn corresponds to the sum of the first $n-2$ integers.

So the total number of terms of our expansion is $$3 + 3 (n-1)+1/2 \, (n-1)(n-2 ) $$ Simplifying and equalizing it to $28$, we get $ 1/2 \, n^2+3/2 \, n+1 =28$, whose positive solution is $n=6$.

Once established the value of $n $ which gives $28$ terms, we can directly calculate the sum of all binomial coefficients of the expansion (all considered as positive) by setting $a=b=c=1$ (or equivalently $x=-2$ in the expression reported in the OP): in fact, in this way any binomial coefficient remains unaltered in the expansion. So we are left with $(1+1+1)^6=3^6=729$, which may potentially be the correct answer in the original interpretation of the person who conceived the problem.

This possible solution is only a hypothesis, driven by the fact that other interpretations seem to give results that are not included in the four possible answers.


As already indicated in the comments there seems to be a typo in the stated problem.

Let \begin{align*} A(x)=\left(1-\frac{2}{x}+\frac{4}{x^2}\right)^n=\sum_{k=-2n}^0a_kx^k\qquad\qquad\qquad n\in\mathbb{N} \tag{1} \end{align*}

We assume $n\in\mathbb{N}$. Since $A(x)$ is the product of $n$ factors $\left(1-\frac{2}{x}+\frac{4}{x^2}\right)$ and each of them contributes to the exponent $k$ of $x^k$ either $0,-1$ or $-2$ the smallest exponent is $-2n$ while the largest is $0$. This explains the index range of the series representation in (1).

According to the problem the sum of the coefficients of $A(x)$ is one of \begin{align*} 64,243,729,2187\tag{2} \end{align*}

The sum of the coeffcients of $A(x)$ is the expression evaluated at $x=1$ and we obtain \begin{align*} \sum_{k=-2n}^0a_k=A(1)=(1-2+4)^n=3^n \end{align*}

Since the result is of the form $3^n$ we can exclude $64=2^6$ leaving three candidates \begin{align*} 243&=3^5\\ 729&=3^6\\ 2187&=3^7 \end{align*} as possible solution with $n=5,6$ or $n=7$.

The index in the series expression of $A(x)$ starts from $-2n$ and ends with $0$. So, the series representation of $A(x)$ consists of at most $2n+1$ different terms.

Substituting the possible candidates $3^5,3^6$ and $3^7$ gives at most $11,13$ and $15$ different terms. In fact, Wolfram Alpha shows they contain precisely $11,13$ resp. $15$ terms.

Conclusion: None of the candidates has a series representation with $28$ different terms and coefficients which sum up to $3^n$. This strongly indicates the problem is not correctly specified.

  • $\begingroup$ Maybe some terms cancel out $\endgroup$
    – J. Deff
    Oct 19, 2016 at 13:48
  • $\begingroup$ @J.Deff: In fact no terms cancel out when $n=5,6$ or $n=7$ as WA shows (see paragraph before conclusion). Since $A(1)=3^n$ there are no other plausible candidates other than $3^n$. Do you agree? $\endgroup$ Oct 19, 2016 at 13:57

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .