Coefficient in Expansion

Note: This solution is long but very explanatory.

On applying the whole square inside the brackets, we get: $[(1+x)(1+3x^2)(1+9x^9)(1+27x^{27})]^2$ $=(1+2x+x^2)(1+9x^6+6x^3)(1+18x^9+81x^{18})(1+54x^{27}+729^{54})$ Now, we want to know the coefficient of $x^{33}$ in the expansion of the above expression, without expanding it. We know that on expanding, the different powers of $x$ would multiply, and the result would be coefficient $\times x^{sum}$ , where sum $=$ sum of multiplying powers. For instance, one of the terms in the above expansion would be $6x^3 \times 18x^9 = 6 \times18 \times x^{9+3} = 108x^{12}$ . Therefore, in order to obtain a power of $33$ , that is, $x^{33}$ , we would want that the sum of the multiplying powers of $x$ should be $33$ . So, noticing the expression we obtained after squaring, the different powers of $x$ , in order of the brackets they are present, are $(0, 1, 2)(0,3,6)(0, 9, 18)(0,27,54)$ . Now, we would try to make a sum of $33$ , using just one power from each bracket. Note that we cannot try to sum powers of same bracket, as their terms are not been multiplied, but added. We know that $54$ is useless to us, as it can never sum to $33$ . Applying some logic, we get that the quadruple $(0,6,0,27)$ adds up to $33$ . We can further check for more quadruples, like $(2,3,0,27)$ or $(2,6,18,0)$ or $(0,0,9,27)$ , but they do not add up to $33$ . So, we can draw the conclusion that only one quadruple, that is $(0,6,0,27)$ produces the power $33$ . Hence, we can recognize the term containing $x^{33}$ as $(1 \times9 \times1 \times 54) \times x^{0+6+0+27} = 486x^{33}$ , and hence obtaining the coefficient of $x^{33}$ as $\fbox{486}$ .

Let us expand the given expression partially:

$\begin{aligned} &\left[(1+x)(1+3x^{3})(1+9x^{9})(1+27x^{27})\right]^{2} \\ &= (1+x)^{2} \cdot (1+3x^{3})^{2} \cdot (1+9x^{9})^{2} \cdot (1+27x^{27})^{2} \\ &= (1+2x+x^{2})(1+6x^{3}+9x^{6})(1+18x^{9}+81x^{18})(1+54x^{27}+729x^{54})\end{aligned}$

After this, it is mainly a matter of intuition and common sense. Looking at the coefficients of the terms, it is obvious that only one multiplication "path", from $1$ to $9x^{6}$ to $1$ and $54x^{27}$ will give us a term with $x^{33}$ . The coefficient of this term is of course $9 \times 54 = \boxed{486}$

Okay, my solution isn't too formal, but it gets the job done. We can rewrite the expression as follows:

$(x+1)^{2} \times (3x^{3}+1)^{2} \times (9x^{9}+1)^{2} \times (27x^{27}+1)^{2}$

Which can further be simplified as:

$(x^{2}+2x+1) \times (9x^{6}+6x{3}+1) \times (81x^{18}+18x{9}+1) \times (729x^{54}+54x^{27}+1)$

Now, we realize that in order to produce a term with an $x^{33}$ , we need to have multiplied together specific terms from this expression. We do not need to multiply the entire expression out to figured out the desired quantity, as that would be quite tedious. Since all of the terms in the expression are distinct, their product will also contain distinct terms, including the term $x^{33}$ .

We find that the product, $9x^{6} \times 54x^{27}$ is the only way for us to get a term of this form.

$9x^{6} \times 54x^{27} = 486x^{33}$

Therefore, the coefficient of $x^{33}$ is $\boxed{486}$ .

First, let's write the expression as $(1+x)^2(1+3x^3)^2(1+9x^9)^2(1+27x^{27})^2$ $=(1+2x+x^2)(1+6x^3+9x^6)(1+18x^9+81x^{18})(1+54x^{27}+27^2x^{54})$ Now if we were to expand this expression, each term would be the product of one term from each of the four parenthesis (e.g. a certain term would be $2x \cdot 9x^6 \cdot 1 \cdot 1$ ). So to find the coefficient of $x^{33}$ , we consider each possible product of four terms which has a resultant $x^{33}$ term. I thought that working backwards, i.e. starting from the fourth parenthesis, would be the easiest. It is obvious that using the $x^{54}$ wouldn't get us anywhere, as $54$ is already above $33$ . So next we try $x^{27}$ (note: i'm only writing the degree right now as we can worry about the coefficients later). We can't take $x^9 \text{ or } x^{18}$ here as both would result in a sum over $33$ . So we choose 1. Now, we must choose $x^6$ and $1$ as choosing $x^3$ would not give any working combinations. So $1, x^6, 1, x^{27}$ works. Now we try using $1$ from the fourth parenthesis to start off, but the largest possible degree we can end with then is $18+6+2 = 26 < 33$ . So the only possible combination is the one noted above, resulting in a coefficient of $9 \cdot 54 = \boxed{486}$

Problem could be expressed as (1+3^{n}x^{3^{n})^{2} from n=0 to 3

$1 + 2(3^{n}x^{3^{n}})+ 3^{2n}x^{2 \times 3^{n}} from n=0 to 3$

n=0, $1 + 2x + x^{2}$

n=1, $1+6x^{3} + 9x^{6}$

n=2, $1+18x^{9} + 81x^{18}$

n=3, $1+54x^27 +729x^{54}$

We know that to get $x^{33}$ we need two powers of x that add up to 33 then multiply their coefficients.

$54x^{27} \times 9x^{6} = \boxed{486x^{33}}$

Desenvolvendo a expressão,

$\\ \left [ \left ( 1 + x \right ) \left ( 1 + 3x^3 \right ) \left ( 1 + 9x^9 \right ) \left ( 1 + 27x^{27} \right )\right ]^2 = \\\\ \left ( 1 + x \right )^2 \cdot \left ( 1 + 3x^3 \right )^2 \cdot \left ( 1 + 9x^9 \right )^2 \cdot \left ( 1 + 27x^{27} \right )^2 = \\\\ \left ( 1 + 2x + x^2 \right ) \left ( 1 + 6x^3 + 9x^6 \right ) \left ( 1 + 18x^9 + 81x^{18} \right ) \left ( 1 + 54x^{27} + 729^{54} \right ) =$

Note que, ao somar os expoentes 0, 6, 0 e 27; respectivamente, obtemos o referido valor em questão!

Logo,

$\\ 1 \cdot 9x^6 \cdot 1 \cdot 54x^{27} = \\\\ \boxed{\boxed{486}x^{33}}$

Let's rewrite the expression a bit differently - Our given equation is the same as -

$(1+2x+x^2)(1+6x^3+9x^6)(1+18x^9+81x^{18})(1+54x^{27}+729x^{54})$

Now, if we expand the expression, power of any of it's term will be the sum of a+b+c+d, where a, b, c, d belong to sets A = {0, 1, 2}, B = {0, 3, 6}, C = {0, 9, 18}, D = {0, 27, 54} respectively. The only possible way that happens is when a = 0, b = 6, c = 0, d = 27.

Therefore value of the coefficient will be 54*9 = 486.

This question seems rather complicated but has fairly simple solution. First you need to simplify the polynomials and numbers in the parentheses. You should get an answer now of:

$[ 1 + (27x^9 * x^9 * x^9) + 9x^9 + (243x^9 * x^9 * x^9 * x^9) + 3x^3 + (81x^9 * x^9 * x^9 * x^3) + (27x^9 * x^3) + (729x^9 * x^9 * x^9 * x^9 * x^3) + x + (27x^9 * x^9 * x^9 * x) + (9x^9 * x^9) + (243x^9 * x^9 * x^9 x^9 *x) +3x^4 + (81x^9 * x^9 * x^9 * x^4) + (27x^9 * x^4) + (729x^9 * x^9 * x^9 * x^9 * x^4)]^2$

Now, all you need to do is find ordered pairs $(a,b)$ of indices that have a sum of 33. There are only 2 pairs, $((81x^9 * x^9 * x^9 * x^3), 3x^3)$ and $(3x^3 , (81x^9 * x^9 * x^9 * x^3))$ . The sum of the products of the integers in the pairs is 486. It is easier than squaring the whole equation. Note that I have written certain polynomials in brats as I could not write them in LaTeX.