First slide

Binomial theorem for negative integral and rational Index

Question

The coefficient of x²⁰ in the expansion of ${(1 + x^{2})}^{40} \cdot {(x^{2} + 2 + \frac{1}{x^{2}})}^{- 5}$ is

Moderate

A

$^{30} C_{10}$
B

$^{30} C_{25}$
C

1
D

None of these

Solution

Given,
$\begin{matrix} {(1 + x^{2})}^{40} \cdot {(x^{2} + 2 + \frac{1}{x^{2}})}^{- 5} \\ = {(1 + x^{2})}^{40} {\{{(x + \frac{1}{x})}^{2}\}}^{- 5} \\ = {(1 + x^{2})}^{40} {(x + \frac{1}{x})}^{- 10} \\ = x^{10} {(1 + x^{2})}^{40} {(1 + x^{2})}^{- 10} = x^{10} {(1 + x^{2})}^{30} \end{matrix}$
to find The coefficient of x²⁰ in the expansion of
${(1 + x^{2})}^{20} {(x + \frac{1}{x})}^{- 10} i.e. x^{10} {(1 + x^{2})}^{30}$
Now, ${(1 + x^{2})}^{30} =^{30} C_{0} +^{30} C_{1} (x^{2}) +^{30} C_{2} {(x^{2})}^{2} +^{30} C_{3} {(x^{2})}^{3} +^{30} C_{4} {(x^{2})}^{4} + \dots +^{30} C_{30} {(x^{2})}^{30} \dots (i)$
$x^{10} {(1 + x^{2})}^{30} = x^{10} +^{30} C_{1} x^{12} +^{30} C_{2} x^{14} +^{30} C_{3} x^{16} +^{30} C_{4} x^{18} +^{30} C_{5} x^{20} + \dots + x^{70}$
$∴ Coefficient of x^{20} {is}^{30} C_{5} {or}^{30} C_{25} .$ $[∵^{n} C_{r} =^{n} C_{n - r}]$

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