Find the 11th term from the beginning and the 11th term from the end in the expansion of $(2 x - {\frac{1}{x^{2}}}^{25})$

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Solution

$T_{r + 1} = T_{n} = (- 1)^{r}^{n} C_{r} x^{n - r} y^{r}$

$T_{11} = T_{10 + 1} = (- 1)^{10}^{25} C_{10} (2 x)^{15} {(\frac{1}{x^{2}})}^{10}$

$=^{25} C_{10} (\frac{2^{15}}{x^{5}}) = \frac{25!}{10! 5!} 2^{15} x^{15} \times x^{- 20}$

11th term from the end $= (26 - 11 + 1) = 16^{t h}$ from beginning.

$\Rightarrow T_{16} = T_{15 + 1}$

$= (- 1)^{15}^{25} C_{15} (2 x)^{10} {(\frac{1}{x^{2}})}^{15} =^{- 25} C_{15} \frac{2^{10}}{x^{20}}$

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