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!
 A: 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 $a$.
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.
A: 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 $a$.  
