number of terms in the expansion containing powers of $x$

Question

How do i find the number of terms containing powers of $x$ in the expansion of: $$(1+x)^{100}(1+x^2-x)^{101}$$

I tried using $(1+x)((1+x(1+(x)^2-x))^{100})$ which simplified into : $$(1+x)(1+x^3)^{100}$$ but i'm not sure if this is the correct approach and also what do i do to simplify it further to get the answer?

It's a multiple choice question with options A:202, B:302, C:301 and D:101 please explain the method to solve questions of this type :)

Answer 1

Hint: note that $(a+(x^{m}))^{n} = \sum_{k=0}^{n} {{n}\choose{k}}a^{n-k}(x^{m})^{k}.$

Answer 2

Yes you are correct and now first expansion is having 2 terms while second is having 101 terms so total number of terms are 202 ...

Answer 3

There are 101 terms in the first factor, of which 100 have a power of $x$ greater than 0. The second factor can be written:

$$ \sum_{k_1 + k_2 + k_3 = 101} 1^{k_1} \cdot x^{2 k_2} \cdot x^{k_3} $$

(just imagine the product multiplied out) so you are asking how many sets of (k_1, k_2, k_3), all integers at least 0, are such that $k_1 + k_2 + k_3 = 101$ with $k_1 < 101$ (that one gives the term 1, there is just one of those).

Let's find out how many solutions $k_1 + k_2 + k_3 = 101$ has, this is like chopping a line of 101 $*$ into three pieces, say by separating with $|$ (this is called a stars and bars argument, for obvious reasons). But then you have a total of $101 + 2$ positions to be filled with 101 stars and 2 bars, that can be done in $\binom{101 + 2}{2} = 5253$ ways, of which you subtract 1 for $k_1 = 101$.

Combining your factors, you have 101 terms in the first factor, $5252$ with $x$ from the second, and the terms with $x$ in the product are the result of multiplying any term from the first factor with a term containing $x$ from the second, i.e., $101 \cdot 5252 = 53042$ (as long as no simpolifications happen).

A lower bound is that the result is a polynomial of degree $100 + 2 \cdot 101 = 302$, so there are at most $301$ terms with powers of $x$. But there are negative terms, so cancellation can/will happen, and you get less.