Animated Solution for Mathematics - Binomial Theorem: The coefficient of $x^7$ in $(1-x+2x^3{)}^{10}$ is —————————— .

Welcome, future engineers. Today, we are going to dismantle a problem that often intimidates students simply because it looks "too big." We are looking for the coefficient of $x^7$ in the expansion of $(1-x+2x^3{)}^{10}$.

When you see three terms inside a bracket raised to a power, your first instinct might be to panic or try to force a binomial expansion. But I want you to take a deep breath. We are going to use the Multinomial Theorem—the "big brother" of the binomial theorem—to solve this with elegance and precision.

The DNA of the Expansion

Every term in this expansion is a unique combination of our three base terms: $1$, $-x$, and $2x^3$. If we choose the first term $a$ times, the second term $b$ times, and the third term $c$ times, the total number of choices must equal the power of the expansion, which is $10$.

This gives us our first fundamental constraint: $a+b+c=10$.

The general term $T$ is given by the multinomial coefficient multiplied by the terms raised to their respective powers:

Let us simplify this. Since $1^a$ is always $1$, it disappears from our calculation. We are left with $(-x{)}^b$ and $(2x^3{)}^c$.

Grouping these, we get:

This is the "DNA" of our expansion. The coefficient part is $\frac{10!}{a!b!c!}(-1{)}^b(2{)}^c$, and the power of $x$ is determined entirely by $b+3c$.

The Rules of the Game

We are hunting for $x^7$. This means our power constraint is $b+3c=7$. Combined with our sum constraint $a+b+c=10$, we have a system of two equations with three variables.

Since $a,b,c$ must be non-negative integers, we have a finite, manageable search space. Let us test values for $c$, as it is the most constrained variable.

If $c=0$, our power constraint becomes $b+3(0)=7$, so $b=7$. Plugging these into the sum constraint, $a+7+0=10$, we find $a=3$.

All are non-negative integers! The coefficient is:

If $c=1$, then $b+3(1)=7$, which gives $b=4$. The sum constraint $a+4+1=10$ gives $a=5$. This is a valid set.

The coefficient is:

If $c=2$, then $b+3(2)=7$, so $b=1$. The sum constraint $a+1+2=10$ gives $a=7$. Another valid set!

The coefficient is:

The Final Summation

What about $c=3$? If we test it, $b+3(3)=7$ leads to $b=-2$. Since we cannot have a negative number of selections, this case is impossible.

We have exhausted all possibilities. Now, we simply add the results of our valid cases to find the total coefficient of $x^7$:

Calculating this, we get $2520-1560=960$.

And there you have it. By systematically breaking down the problem into constraints and cases, we turned a daunting expression into a simple arithmetic sum. Keep this methodical approach in your toolkit, and no JEE problem will ever be too big for you. The final answer is 960.