Answer:
[tex]256+\dfrac{1024x}{3}+\dfrac{1792x^2}{9}+\dfrac{1792x^3}{27}[/tex]
Step-by-step explanation:
You want the first 4 terms in the expansion of the binomial power ...
(2 +x/3)⁸
In general, the expansion of binomial (p +q) to the power n will have the form ...
[tex](p+q)^n=a_0p^nq^0+a_1p^{n-1}q^1+\dots+a_kp^{n-k}q^k+\dots+a_np^0q^n[/tex]
In this expansion, the coefficients a₀, a₁, a₂, ..., aₙ are called the binomial coefficients. They are computed using the formula ...
[tex]a_k=nCk=\dfrac{n!}{k!(n-k)!}[/tex]
where nCk is "n choose k", the number of ways k objects can be chosen from n objects.
Pascal's triangle, a partial version of which is shown in the attachment, is a visual representation of the values of nCk for different n and k. The numbering starts from 0 in both rows and diagonals. Row 8 is the last row shown here. That row gives the set of coefficients we need for the 8-th power expansion in this problem. It tells you ...
Using these relations for the application at hand, we have ...
p = 2
q = x/3
So, the first 4 terms of the expansion are ...
[tex]\left(2+\dfrac{x}{3}\right)^8=1\cdot2^8+8\cdot2^7\left(\dfrac{x}{3}\right)^1+28\cdot2^6\left(\dfrac{x}{3}\right)^2+56\cdot2^5\left(\dfrac{x}{3}\right)^3+\dots\\\\\\=\boxed{256+\dfrac{1024x}{3}+\dfrac{1792x^2}{9}+\dfrac{1792x^3}{27}+\dots}[/tex]
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Additional comment
Pascal's triangle has several interesting properties. The number in each row is the sum of the two numbers immediately above. If there is no number above, it is taken to be zero. That makes the left and right sides of the triangle be ones. The first diagonal is the counting numbers, and the second diagonal is the set of triangular numbers (the cumulative sums of the counting numbers). The triangle is symmetric.
