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

Question

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!

Answer 1

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.

Answer 2

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 $a$.

Answer 3

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 $a$.