Finding the Coefficient for $a^4$ in a Series with Variable Exponents

[anemone]

What is the coefficient for $a^4$ in $(1-a^2)+(1-a^2)^2++(1-a^2)^3+\cdots+(1-a^2)^{12}$?

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[anemone]

Congratulations to the following members for their correct solutions::)

1. chisigma
2. SuperSonic4
3. greg1313
4. lfdahl
5. kaliprasad

Solution from SuperSonic4:

Spoiler

The binomial theorem states that $$\sum_{k=0}^{n}(1+x)^n = {n \choose k} x^k$$In this case $$x = -a^2$$ and so we have $$\sum_{k=0}^{n}(1-a^2)^n = {n \choose k} (-a^2)^k = {n \choose k} (-a)^{2k}$$

As we are dealing with integer powers it follows that $$2k$$ is always even and so our sign will always be plus.

For the exponent to be 4 $$2k = 4 \therefore k=2$$

The coefficient of any given power is [math]{n \choose k}[/math] and since we've just worked out that [math]k=2[/math] we can work out the coefficient of [math]x[/math] for each term and using inspection we can clearly see that the coefficient of [math]x^4[/math] in the first term is [math]0[/math].

Thus we start with
$${2 \choose 2} = 1$$ and work our way sequentially up to $${12 \choose 2} = 66 $$ using good old fashioned brute force

The coefficients for each term are 0, 1, 3, 6, 10, 15, 21, 28, 36, 45, 55, 66 which also happen to be the triangular numbers

Thus the total coefficient is all these added together: [math]286[/math]

Solution from lfdahl:

Spoiler

\[(1-a^2)+(1-a^2)^2+ ... + (1-a^2)^{12} \\\\= \frac{1-(1-a^2)^{13}{}}{1-(1-a^2)}-1 =\frac{1-\sum_{j=0}^{13}\binom{13}{j}(-1)^{j} a^{2j}}{a^2}-1 \\\\ =\sum_{j=1}^{13}\binom{13}{j}(-1)^{j+1} a^{2(j-1)}-1\]
Thus the coefficient of $a^4$ (for $j=3$) is: $\binom{13}{3} = 286$.

Solution from kaliprasad:

Spoiler

If we put $x= (1-a^2)$

We get given expression as

$x+x^2+x^3+x^4+x^5+x^6+x^7+x^8+x^9+x^{10}+x^{11}+x^{12}$

= $x\dfrac{1-x^{12}}{1-x}$

= $(1-a^2)\dfrac{1-(1-a^2)^{12}}{a^2}$

= $\dfrac{(1-a^2)-(1-a^2)^{13}}{a^2}$

Now $\dfrac{1-a^2}{a^2}$ shall not contribute to $a^4$ so we need to find the coefficient of $a^6$ in $-(1-a^2)^{13}$ which is ${13\choose 3}$

Hence the ans is ${13\choose 3}$

Alternative solution:

Spoiler

Notice that

1. The first binomial which raised to the first power, $(1-x^2)^1$ has no term of $x^4$.

2. The subsequent binomials, which raised to the second power all the way up to the twelve power, $x^4$ lies on the third term (and is a positive term) if we are to expand them out and writing the result in ascending powers of $x$.

http://mathhelpboards.com/attachment.php?attachmentid=3651&stc=1
Hockey Stick Pattern tells us the coefficient for $a^4$ in $(1-a^2)+(1-a^2)^2++(1-a^2)^3+\cdots+(1-a^2)^{12}$ is 286.