Animated Solution for Mathematics - Binomial Theorem: Let $[t]$ denote the greatest integer $\le t$. if the constant term in the expansion of $(3x^2-\frac{1}{2x^5}{)}^7$ is $\alpha$ then $[\alpha ]$ is equal to ——————

Welcome, fellow traveler of the mathematical landscape. Today, we are going to dissect a classic problem that often appears in the JEE Advanced arena.

It is a problem that tests your precision, your ability to manipulate algebraic structures, and your patience with arithmetic. We are looking for the 'constant term' in the expansion of $(3x^2-\frac{1}{2x^5}{)}^7$.

Imagine this expression as a vast, unfolding structure. When we expand it, we get a series of terms, and somewhere in that series, there is a term that stands alone, independent of $x$. That is our target, $\alpha$.

Our Telescope

We do not have the time or the inclination to expand the entire binomial expression. That would be a tedious, error-prone endeavor. Instead, we use the Binomial Theorem's most powerful tool: the general term formula.

For any expansion of $(a+b{)}^n$, the $(r+1)$-th term is given by:

This formula is our telescope. It allows us to zoom in on any specific term without having to look at the rest of the expansion. In our problem, we identify $n=7$, $a=3x^2$, and $b=-\frac{1}{2x^5}$.

Notice the negative sign attached to the second term; ignoring it is the most common trap students fall into. Let us substitute these values into our formula:

Now, we need to separate the constants from the variables. This is where the beauty of algebra shines. We rearrange the expression to group the coefficients and the powers of $x$ separately:

By applying the laws of exponents, we simplify the powers of $x$. When we raise a power to a power, we multiply the exponents. Thus, $(x^2{)}^{7-r}$ becomes $x^{14-2r}$, and $(x^{-5}{)}^r$ becomes $x^{-5r}$.

When we multiply these, we add the exponents:

We are looking for the constant term. By definition, a constant term has no $x$ component. This means the power of $x$ must be zero.

We set our exponent to zero:

Solving this is straightforward: $7r=14$, which gives us $r=2$. This is the key that unlocks the door. We now know that the constant term is the third term ($T_{2+1}=T_3$) of the expansion.

With $r=2$ in hand, we return to our expression to calculate $\alpha$:

Now, we compute the components. The combination ${}^7{C}_2$ is:

The power $3^5$ is $243$. Putting it all together:

Performing the division, we get $\alpha =1275.75$. Finally, we apply the greatest integer function, $[\alpha ]$.

This function simply asks for the largest integer less than or equal to our value. For $1275.75$, that is clearly $1275$. You have successfully navigated the binomial maze.