Question

What is the seventh term of (x + 4)8?

Answer 1

The (p+1)-th term of the Newton binomial expansion

$(a+b)^{n}$

is given by

$t_{p+1}=\dbinom{n}{p}\,a^{n-p}\,b^{p}$
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We want the 7th term. Hence, we set $p+1$ to be $7:$

$p+1=7~~\Rightarrow~~p=6$


Then, the 7th term is

$t_{7}=\dbinom{8}{6}\,x^{8-6}\,4^{6}\\\\\\ t_{7}=\dfrac{8!}{6!\cdot (8-6)!}\cdot x^{8-6}\cdot 4^{6}\\\\\\ t_{7}=\dfrac{8\cdot 7\cdot \diagup\!\!\!\! 6!}{\diagup\!\!\!\! 6!\cdot 2!}\cdot x^{2}\cdot 4^{6}\\\\\\ t_{7}=\dfrac{8\cdot 7}{2\cdot 1}\cdot x^{2}\cdot 4^{6}\\\\\\ t_{7}=28\cdot x^{2}\cdot 4,096\\\\ \boxed{\begin{array}{c} t_{7}=114,688\,x^{2} \end{array}}$


Correct answer: $\text{A. }114,688\,x^{2}.$