Is there a way to find the specific variable coefficient in a binomial expansion?

If a problem asks to find the coefficient of a variable, say, $$x^2$$, in a large binomial expansion, is there a way to solve without doing the whole expansion (I do it with Pascal's Triangle / Binomial Theorem). For example, in this problem

The coefficient of $$x^2$$ in the expansion of $$(\frac{1}{x} + 5x)^8$$ is equal to the coefficient of $$x^4$$ in the expansion of $$(a+5x)^7$$, $$a$$ is a real number. Find the value of $$a$$.

I expand it out and get different answers on different tries. Not sure what's the best method to proceed. If anyone could help I would appreciate it so much!

• For some basic information about writing mathematics at this site see, e.g., here, here, here and here. Aug 2, 2019 at 11:29

The binomial theorem tells you that $$\left(\frac1x + 5x\right)^8 = \sum_{i = 0}^8\binom8i\frac{1}{x^i}(5x)^{8-i}\\ (a + 5x)^7 = \sum_{j = 0}^7\binom7ja^j(5x)^{7-j}$$ Since we're looking for the $$x^2$$ term in the first sum, that happens only when $$i = 3$$. For the second sum we're interested in the $$x^4$$ term which only is when $$j = 3$$. We get $$\binom83\frac1{x^3}(5x)^5 = 56\cdot 5^5x^2\\ \binom73a^3\cdot(5x)^4 = 35a^3\cdot 5^4x^4$$ Now equate the two coefficients, and solve for $$a$$.

It's easier if you find the coefficient of $$x^{10}$$ in $$(1+5x^2)^8$$, which is the same as the coefficient of $$x^5$$ in $$(1+5x)^8$$; this is $$\binom{8}{5}\cdot 5^5=\binom{8}{3}\cdot 5^5$$ The coefficient of $$x^4$$ in the expansion of $$(a+5x)^7$$ is $$\binom{7}{4}\cdot a^3\cdot 5^4=\binom{7}{3}\cdot a^3\cdot 5^4$$ Now the equation is easy.

This is exactly what the binomial formula is for!

$$(a+b)^n = \sum_{i=0}^n {n \choose i} a^ib^{n-i}$$

I'll show you how to do the first one. Start by plugging in $$a= 1/x$$ and $$b = 5x$$ and $$n=8$$ and simplify:

$$(1/x+5x)^8 = \sum_{i=0}^8 {8 \choose i} (1/x)^i(5x)^{8-i} = \sum_{i=0}^8 {8 \choose i} 5^{8-i} \frac{x^{8-i}}{x^i} = \sum_{i=0}^8 {8 \choose i} 5^{8-i} x^{8-2i}.$$

This is the sum of nine terms, one for each $$i=0,1,\ldots, 8$$. Notice each term has a different power of $$x$$. So the coefficient of $$x^2$$ happens when $$n-i = 2$$ which is when $$i=3$$. That means the coefficient is

$${8 \choose 3} 5^{8-3} = {8 \choose 3} 5^{5}$$

which you can simplify.

Do the same for the other binomial and equate the two answers and then solve for $$a$$.