Question

Using the binomial theorem to expand (1+4x)^4​

Answer 1

Answer:

${256x}^{4}+{256x}^{3} + {96x}^{2} + 16x + 1$

Explanation:

${(1 + 4x)}^{4} = (1 + 4x)(1 + 4x)(1 + 4x)(1 + 4x)$

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 $\binom{n}{k} = \frac{n!}{k!(n-k)!}$

Which, of course, we have to multiply for the product of the two terms.

For example, the coefficient of the third grade term in ${(1+4x)}^{4}$ is $\binom{4}{3}=4$

So we have ${(1 + 4x)}^{4} = \binom{4}{4} \cdot {(4x)}^{4}+\binom{4}{3} \cdot {(4x)}^{3} + \binom{4}{2} \cdot {(4x)}^{2} + \binom{4}{1} \cdot 4x + \binom{4}{0} \cdot 1$

Which is equal to $1 \cdot {(4x)}^{4}+4 \cdot {(4x)}^{3} + 6 \cdot {(4x)}^{2} + 4 \cdot 4x + 1 \cdot 1 = {256x}^{4}+{256x}^{3} + {96x}^{2} + 16x + 1$

Hope this helps :)