To solve this problem, we use the Binomial Theorem to compute the expansion:
[tex](a+b)^n=\sum_{k=0}^n\binom{n}{k}a^{n-k}b^k=\sum_{k=0}^n\frac{\text{ n!}}{(n-k)!k!}a^{n-k}b^k.[/tex]
Therefore, the first term of the expansion is:
[tex]\frac{8!}{(8-0)!0!}x^{8-0}2^0=\frac{8!}{8!1}x^8*1=x^8.[/tex]
Using the binomial theorem, we get that:
[tex](x+2)^8=x^8+16x^7+112x^6+448x^5+1120x^4+1792x^3+1792x^2+1024x+256.[/tex]
Therefore, the first three terms are:
[tex]x^8,16x^7,112x^6.[/tex]
Answer:
[tex]x^{8}, 16x^{7}, 112x^{6}[/tex]
