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प्रश्न
Find the 11th term from the beginning and the 11th term from the end in the expansion of \[\left( 2x - \frac{1}{x^2} \right)^{25}\] .
उत्तर
Given: \[\left( 2x - \frac{1}{x^2} \right)^{25}\] Clearly, the given expression contains 26 terms.
So, the 11th term from the end is the (26 − 11 + 1)th term from the beginning. In other words, the 11th term from the end is the 16th term from the beginning.
Thus, we have:
\[T_{16} = T_{15 + 1} =^{25}{}{C}_{15} (2x )^{25 - 15} \left( \frac{- 1}{x^2} \right)^{15} \]
\[ = ^{25}{}{C}_{15} \left( 2^{10} \right)\left( x^{10} \right)\left( \frac{- 1}{x^{30}} \right) = - ^{25}{}{C}_{15} \left( \frac{2^{10}}{x^{20}} \right)\]
Now, we will find the 11th term from the beginning.
\[T_{11} = T_{10 + 1} \]
\[ = ^{25}{}{C}_{10} (2x )^{25 - 10} \left( \frac{- 1}{x^2} \right)^{10} \]
\[ = ^{25}{}{C}_{10} \left( 2^{15} \right)\left( x^{15} \right)\left( \frac{1}{x^{20}} \right)\]
\[ = ^{25}{}{C}_{10} \left( \frac{2^{15}}{x^5} \right)\]
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