# 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 ?

You got the first $$3$$ terms of $$(2-3x)^6$$ correct: $$64-576x+2160x^2$$.

Now multiply:

$$\left(1+\dfrac x2\right)(64-576x+2160x^2...)=(64-576x+2160x^2...)+(32x-288x^2...)$$

$$=64-544x+1872x^2...$$

Add the two together, and keep the terms with $$x^0$$, $$x^1$$, and $$x^2$$. So the answer is $$64+(32-576)x+(2160-288)x^2$$

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. $$(1+x/2)(2-3x)^6= 64 - 544 x + 1872 x^2 - 3240 x^3 + 2700 x^4 - 486 x^5 - 729 x^6 + (729 x^7)/2$$