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Q.

Let [t] denote the greatest integer <t. If the constant term in the expansion of (3x212x5)7  is  α,  then  [α]  is equal to …… 

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answer is 1275.

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Detailed Solution

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 General term Tr+1=7Cr3x27r12x5r
Tr+1=(1)r7Cr37r2rx147r
14-7r=0r=2
 Constant term =a=(1)27C23522=51034
a=1275.75 [a]=1275

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Let [t] denote the greatest integer <t. If the constant term in the expansion of (3x2−12x5)7  is  α,  then  [α]  is equal to ……