Animated Solution for Mathematics - Binomial Theorem: The coefficient of $x^7$ in the expansion of $(1-x-x^2+x^3{)}^6$ is

Imagine you are staring at the expression $(1-x-x^2+x^3{)}^6$. It looks like a monster, but in the world of JEE Advanced, monsters are often just illusions.

Look at the inner polynomial: $1-x-x^2+x^3$. If we group the first two terms as $(1-x)$ and the last two as $-x^2(1-x)$, we see a common factor.

We can rewrite the base as $(1-x)(1-x^2)$. Now, our expression becomes:

We have successfully transformed a nightmare into a product of two standard binomials.

Now that we have $(1-x{)}^6(1-x^2{)}^6$, we need the coefficient of $x^7$. We treat these two binomials as independent entities.

For the first part, $(1-x{)}^6$, the general term is:

For the second part, $(1-x^2{)}^6$, we use a different index, $k$. Its general term is:

When we multiply these together, we get the combined general term:

We are hunting for the coefficient of $x^7$. This means the exponent of $x$ in our combined term, $r+2k$, must equal $7$.

We have the Diophantine equation $r+2k=7$, subject to the constraints $0\le r\le 6$ and $0\le k\le 6$. Let's test values for $k$:

If $k=0$, $r=7$ (Invalid, $r>6$). If $k=1$, $r=5$ (Valid). If $k=2$, $r=3$ (Valid). If $k=3$, $r=1$ (Valid). * If $k=4$, $r=-1$ (Invalid).

We calculate the coefficients for each valid pair $(r,k)$:

For $(r=5,k=1)$:

For $(r=3,k=2)$:

For $(r=1,k=3)$:

Summing these up, we get $36-300+120=-144$. The final coefficient is $-144$.