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प्रश्न
Find the 7th term from the end in the expansion of \[\left( 2 x^2 - \frac{3}{2x} \right)^8\] .
उत्तर
Let Tr+1 be the 7th term from the end in the given expression.
Then, we have:
Tr+1 = (9 − 7 + 1) = 3rd term from the beginning
Now,
\[T_3 = T_{2 + 1} \]
\[ = ^{8}{}{C}_2 (2 x^2 )^{8 - 2} \left( - \frac{3}{2x} \right)^2 \]
\[ = \frac{8 \times 7}{2 \times 1}\left( 64 x^{12} \right)\frac{9}{4 x^2}\]
\[ = 4032 x^{10}\]
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