The first 3 terms of the expansion of $\left(1+\frac{x}{2}\right)\left(2-3x\right)^6$ According to the ascending powers of $x$, find the first $3$ terms if the expansion of 
$$\left(1+\frac{x}{2}\right)\left(2-3x\right)^6$$
For the expansion of $$(2-3x)^6 $$
The first 3 terms are $$64 -576x + 2160x^2$$
Now, are the required 3 terms are 
$$64-576x + 2160x^2$$
Or
$$32x -288 x^2 +1080 x^3 $$
Or otherwise ?
 A: You got the first $3$ terms of $(2-3x)^6$ correct:  $64-576x+2160x^2$.
Now multiply:   
$\left(1+\dfrac x2\right)(64-576x+2160x^2...)=(64-576x+2160x^2...)+(32x-288x^2...)$
$=64-544x+1872x^2...$
A: Add the two together, and keep the terms with $x^0$, $x^1$, and $x^2$. So the answer is $$64+(32-576)x+(2160-288)x^2$$
A: The Binomial Theorem gives:
$$
(a+b)^n= \sum_{i=0}^n \binom{n}{i} a^i b^{n-i}
$$
Using this you can find each power you need from the $(2-3x)^6$ term. For instance, the $x$ term from this is $\binom{6}{1} (-3x)^1 2^5= 6 \cdot 2^5 \cdot (-3) \cdot x= -576 x$. You can then multiply this times the constant term from $(1+x/2)$. Then you just need to add that to the $x/2$ times the constant term of $(2-3x)^6$. Proceeding this way gives you all the terms. Once you have the idea, the rest is work, so you can also shortcut/check (more the latter) using this link here will give you the complete expansion near the bottom of the loaded page.
$$
(1+x/2)(2-3x)^6= 64 - 544 x + 1872 x^2 - 3240 x^3 + 2700 x^4 - 486 x^5 - 729 x^6 + (729 x^7)/2
$$
