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TG EAMCET 4 May 2025 Shift 1 Paper

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Question : 20 of 160

Marks: +1, -0

When

\left|x\right| < \frac{1}{2}

the coefficient of

x^6

in the expansion of

\left(\frac{2-x}{1+2x}\right)^2

1320
2640
1088
1980

Solution:

Coefficient of

x^6

in

(x-2)^2(1+2x)^{-2}

= (x^2-4x+4) \left(1-4x+ \frac{-2(-2-1) \times 4 x^2}{2} + .\right)

- \frac{2(-2-1)(-2-2) \times 8 x^3}{6} +

\frac{-2(-2-1)(-2-2)(-2-3) \times 16 x^4}{24} +

- \frac{2(-2-1)(-2-2)(-2-3)(-2-4) \times 32 x^5}{5!} +

- \frac{2(-2-1)(-2-2)(-2-3)(-2-4)(-2-5) \times 64 x^6) }{6!}

So, coefficient of

x^6

equals to

= \frac{16}{24} \times (-2) \times (-3) \times (-4) \times (-5)

\frac{-4 \times 32}{5!} (-2) \times (-3) \times (-4) \times (-5) \times (-6)

+ \frac{4 \times 64}{6!} \times -2 \times (-3) \times -4 \times -5 \times (-6) \times -7

= 2640

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