Answer:
[tex] {256x}^{4}+{256x}^{3} + {96x}^{2} + 16x + 1 [/tex]
Explanation:
[tex] {(1 + 4x)}^{4} = (1 + 4x)(1 + 4x)(1 + 4x)(1 + 4x)[/tex]
The coefficient of the kth term (ordering in increasing order for the exponent of x) is just the number of ways we have to choose k factors from that expression, so if we let k be the exponent of x, and n be the total number of terms, the coefficient of x^k is [tex] \binom{n}{k} = \frac{n!}{k!(n-k)!} [/tex]
Which, of course, we have to multiply for the product of the two terms.
For example, the coefficient of the third grade term in [tex] {(1+4x)}^{4} [/tex] is [tex] \binom{4}{3}=4[/tex]
So we have [tex] {(1 + 4x)}^{4} = \binom{4}{4} \cdot {(4x)}^{4}+\binom{4}{3} \cdot {(4x)}^{3} + \binom{4}{2} \cdot {(4x)}^{2} + \binom{4}{1} \cdot 4x + \binom{4}{0} \cdot 1 [/tex]
Which is equal to [tex] 1 \cdot {(4x)}^{4}+4 \cdot {(4x)}^{3} + 6 \cdot {(4x)}^{2} + 4 \cdot 4x + 1 \cdot 1 = {256x}^{4}+{256x}^{3} + {96x}^{2} + 16x + 1 [/tex]
Hope this helps :)
