Let [t] denote the greatest integer leq t. If | vedclass

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(B) The general term in the expansion of $\left(3x^2 - \frac{1}{2x^5}\right)^7$ is given by $T_{r+1} = {}^7C_r (3x^2)^{7-r} \left(-\frac{1}{2x^5}\righ

Vedclass · Accepted Answer

$1275$

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(B) The general term in the expansion of $\left(3x^2 - \frac{1}{2x^5}\right)^7$ is given by $T_{r+1} = {}^7C_r (3x^2)^{7-r} \left(-\frac{1}{2x^5}\right)^r$.Simplifying the expression,we get $T_{r+1} = {}^7C_r \cdot 3^{7-r} \cdot (-1/2)^r \cdot x^{14-2r-5r} = {}^7C_r \cdot 3^{7-r} \cdot (-1/2)^r \cdot x^{14-7r}$.For the constant term,the power of $x$ must be $0$,so $14 - 7r = 0$,which gives $r = 2$.Substituting $r = 2$ into the expression,we get $\alpha = {}^7C_2 \cdot 3^{7-2} \cdot (-1/2)^2$.$\alpha = 21 \cdot 3^5 \cdot \frac{1}{4} = \frac{21 \cdot 243}{4} = \frac{5103}{4} = 1275.75$.Since $[t]$ denotes the greatest integer $\leq t$,we have $[\alpha] = [1275.75] = 1275$.

Let $[t]$ denote the greatest integer $\leq t$. If the constant term in the expansion of $\left(3x^2 - \frac{1}{2x^5}\right)^7$ is $\alpha$,then $[\alpha]$ is equal to $............$.

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