The coefficient of x20 in the expansion of 1-x240x2-2-1x2 is

x^2(1+x^2)^40+2(1+x^2)^40+1/x^2(1+x^2)^40 writing general tence for all three, x^2× ^40C_r_1(x^2)^r_1,2× ^40C_2× (x^2)^r_2,1/^2× ^40C_3(x^2)^r_3 For ...

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Standard XII

Mathematics

Middle Terms

The coefficie...

Question

The coefficient of $x^{20}$ in the expansion of ${(1 + x^{2})}^{40} (x^{2} + 2 + \frac{1}{x^{2}})$ is

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Solution

$x^{2} (1 + x^{2})^{40} + 2 (1 + x^{2})^{40} + \frac{1}{x^{2}} (1 + x^{2})^{40}$
writing general tence for all three,
$x^{2} \times^{40} C_{r_{1}} (x^{2})^{r_{1}}, 2 \times^{40} C_{2} \times (x^{2})^{r_{2}}, \frac{1}{^{2}} \times^{40} C_{3} (x^{2})^{r_{3}}$
For $x^{20},$
$r_{1} = 9, r_{2} = 10, r_{3} = 11$
$\Rightarrow$ cofficients $=^{40} C_{9} + 2 \times^{40} C_{10} +^{40} C_{11}$
As $^{n} C_{r} +^{n} C_{r - 1} =^{n + 1} C_{r}$
$\Rightarrow$ coeff. $=^{41} C_{10} +^{41} C_{11} =^{42} C_{11}$
$\Rightarrow$ No option

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The coefficient of x20 in the expansion of (1+x2)40(x2+2+1x2) is

x2(1+x2)40+2(1+x2)40+1x2(1+x2)40writing general tence for all three,x2×40Cr1(x2)r1,2×40C2×(x2)r2,12×40C3(x2)r3For x20,r1=9,r2=10,r3=11⇒ cofficients =40C9+2×40C10+40C11As nCr+nCr−1=n+1Cr⇒ coeff. =41C10+41C11=42C11⇒ No option

The coefficient of $x^{20}$ in the expansion of ${(1 + x^{2})}^{40} (x^{2} + 2 + \frac{1}{x^{2}})$ is