What is the coefficient of the x4-term in the binomial expansion of (x + 3)¹²?

Solution:

The given binomial expansion is (x + 3)¹² ------------------- (1)

The general form of binomial expansion is (a + b)ⁿ -------- (2)

Comparing (1) and (2)

a = x

b = 3

n = 12

We have to find the coefficient of the term x⁴

This implies r = 3

The terms in the expansion can be obtained using

\(T_{r+1}=\, ^{n}C_{r}a^{(n-r)}b^{r}\)

Now, \(\\T_{3+1}=\, ^{12}C_{3}(x)^{(12-3)}(3)^{3}\\T_{4}=\, ^{12}C_{3}(x)^{9}(3)^{3}\)

\(T_{4}=\, ^{12}C_{3}x^{9}(27)\)

\(^{12}C_{3}=\frac{12\times 11\times 10\times 9\times 8\times 7\times 6\times 5\times 4\times 3\times 2\times 1}{(3\times 2\times 1)(9\times 8\times 7\times 6\times 5\times 4\times 3\times 2\times 1)}\)

\(^{12}C_{3}=\frac{12\times 11\times 10}{3\times 2\times 1}=\frac{12\times 110}{6}=220\)

So, \(T_{4}=220(x^{9})(27)=5940x^{9}\)

Therefore, the coefficient of the xx4-term is 5940.