I have been doing questions from the past year and I come across this question which stumped me:

The constant term in the expansion of $\left(\frac1{x^2}+ax\right)^6$ is $1215$; find the value of $a$. (The given answer is: $\pm 3$ )

Should be an easy one, but I don't know how to begin. Some help please?

  • $\begingroup$ I'm not sure how you could further expand $\frac1{x^2}+ax$... $\endgroup$ – J. M. is a poor mathematician Sep 6 '11 at 7:47
  • $\begingroup$ @J.M. Sorry! My mistake! I have edited the question. $\endgroup$ – Sophia Sep 6 '11 at 7:55
  • $\begingroup$ In that case: you know what the general term of a binomial expansion looks like, yes? $\endgroup$ – J. M. is a poor mathematician Sep 6 '11 at 7:58

Sofia,Sorry for the delayed response.I was busy with other posts.

you have two choices.One is to use pascals triangle and the other one is to expand using the binimial theorem.

You can compare the expression $$\left ( \frac{1}{x^2} + ax \right )^6$$ with $$(a+x)^6$$ where a = 1/x^2 and x = ax,n=6.Here'e the pascal triangle way of expanding the given expression.All you need to do is to substitute the values of a and x respectively.

$$(a + x)^0 = 1$$

$$(a + x)^1 = a +x a+ x$$

$$(a + x)^2 = (a + x)(a + x) = a^2 + 2ax + x^2$$

$$(a + x)^3 = (a + x)^2(a + x) = a^3 + 3a^2x + 3ax^2 + x^3$$

$$(a + x)^4 = (a + x)^3(a + x) = a^4 + 4a^3x + 6a^2x^2 + 4ax^3 + x^4$$

$$(a + x)^5 = (a + x)^4(a + x) = a^5 + 5a^4x + 10a^3x^2 + 10a^2x^3 + 5ax^4 + x^5$$

$$(a + x)^6 = (a + x)^5(a + x) = a^6 + 6a^5x + 15a^4x^2 + 20a^3x^3 + 15a^2x^4 + 6ax^5 + x^6$$

Here'e the Binomial theorem way of expanding it out.

$$(1+x)^n = 1 + nx + \frac{n(n-1)}{2!}x^2 + \frac{n(n-1)(n-2)}{3!}x^3 + ...$$

using the above theorem you should get

$$a^6x^6 + 6a^5x^3 + 15a^4 + \frac{20a^3}{x^3} + \frac{15a^2}{x^6}+\frac{6a}{x^9}+\frac{1}{x^{12}}$$

You can now substitute the constant term and get the desired answer


Via the binomial theorem we know the $x^0$ factor (the coefficient of which you call the "constant term") is actually $${6 \choose 2}\left(\frac{1}{x^2}\right)^2(ax)^{6-2}.$$ Since 6 choose 2 is equal to $5\cdot6/2=15$, we have the equality $15a^4=1215$. The prime factorization of 1215 is $5\cdot3^5$, so we arrive at the conclusion that $a^4=3^4$, or $a=\pm3$ (discarding imaginary solutions).

  • $\begingroup$ To add to this solution: the general term looks something like $\binom{6}{k}a^{6-k} x^{6-3k}$. Now, find the value of $k$ that zeroes the exponent of $x$... $\endgroup$ – J. M. is a poor mathematician Sep 6 '11 at 8:03

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