Question:
Let \([t]\) denote the greatest integer \(\le t\). If the constant term in the expansion of
\(\left(3x^{2}-\frac{1}{2x^{5}}\right)^{7}\)
is \(\alpha\), then \([\alpha]\) is equal to ____________.
Let \([t]\) denote the greatest integer \(\le t\). If the constant term in the expansion of
\(\left(3x^{2}-\frac{1}{2x^{5}}\right)^{7}\)
is \(\alpha\), then \([\alpha]\) is equal to ____________.
\(\left(3x^{2}-\frac{1}{2x^{5}}\right)^{7}\)
is \(\alpha\), then \([\alpha]\) is equal to ____________.
Show Hint
Find \(r\) from the general term so that the power of \(x\) is \(14-7r=0\), compute \(\alpha\) and take its greatest integer.
Updated On: Nov 2, 2026
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Verified By Collegedunia
Correct Answer: 1275
Solution and Explanation
General term: \(T_{r+1}=\binom{7}{r}(3x^2)^{7-r}\left(-\frac{1}{2x^5}\right)^{r}=\binom{7}{r}3^{7-r}\left(-\frac12\right)^r x^{14-2r-5r}\).
Constant term needs \(14-7r=0\Rightarrow r=2\).
\(\alpha=\binom{7}{2}\,3^{5}\left(\frac12\right)^2=21\cdot243\cdot\frac14=\frac{5103}{4}=1275.75\).
Hence \([\alpha]=1275\).
Answer: 1275
Constant term needs \(14-7r=0\Rightarrow r=2\).
\(\alpha=\binom{7}{2}\,3^{5}\left(\frac12\right)^2=21\cdot243\cdot\frac14=\frac{5103}{4}=1275.75\).
Hence \([\alpha]=1275\).
Answer: 1275
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