If 1+x+x220=a0+a1x+a2x2⋯+a40x40, then answer the following questions.The value of a0+a1+a2+⋯+a19 isThe value of a02−a12+a22−⋯−a192 isThe value of a0+3a1+5a2+⋯+81a40 is

Let, 1+x+x220=∑r=040 arxr  ………(1)Replacing x by 1/x, we get    1+1x+1x220=∑r=040 ar1xror  1+x+x220=∑r=040 arx40−r       ........(2)Since (1) and (2) are same series, coefficient of xr in (1) : coefficient of xr in (2) ⇒ ar=a40−rIn (1), putting x = 1, we get320=a0+a1+a2+⋯+a40=a0+a1+a2+⋯+a19+a20+a21+an+2+⋯+a40=2a0+a1+a2+⋯+a19+a20 ∵ ar=a40−ror  a0+a1+a2+⋯+a19=12320−a20=12910−a20Also,        a0+3a1+5a2+⋯+81a40 =a0+81a40+3a1+79a39+⋯+39a19+43a21+41a20          =82a0+a1+a2+⋯+a19+41a20=41910−a20+41a20=41×320a02−a12+a22−a32+⋯ suggests that we have to multiply the two expansions.Replacing r by - 1/x in (1), we get1−1x+1x220=a0−a1x+a2x2−⋯+a40x40⇒ 1−x+x220=a0x40−a1x39+a2x38−⋯+a40     ......(3)Clearly,           a02−a12+a22+⋯+a402 is the coefficient of x40 in 1+x+x2201−x+x220= Coefficient of x40in⁡1+x2+x420In 1+x2+x420, replace  x2 by y, then the coefficient of y20 in 1+y+y220 is a20. Hence, a02−a12+a22−⋯+a402=a20or  a02−a12+a22−⋯−a192+a202+−a212+⋯+a402=a20or  2a02−a12+a22−⋯−a192+a202=a20or  a02−a12+a22−⋯−a192=a2021−a20 Let, 1+x+x220=∑r=040 arxr  ………(1)Replacing x by 1/x, we get    1+1x+1x220=∑r=040 ar1xror  1+x+x220=∑r=040 arx40−r       ........(2)Since (1) and (2) are same series, coefficient of xr in (1) : coefficient of xr in (2) ⇒ ar=a40−rIn (1), putting x = 1, we get320=a0+a1+a2+⋯+a40=a0+a1+a2+⋯+a19+a20+a21+an+2+⋯+a40=2a0+a1+a2+⋯+a19+a20 ∵ ar=a40−ror  a0+a1+a2+⋯+a19=12320−a20=12910−a20Also,        a0+3a1+5a2+⋯+81a40 =a0+81a40+3a1+79a39+⋯+39a19+43a21+41a20          =82a0+a1+a2+⋯+a19+41a20=41910−a20+41a20=41×320a02−a12+a22−a32+⋯ suggests that we have to multiply the two expansions.Replacing r by - 1/x in (1), we get1−1x+1x220=a0−a1x+a2x2−⋯+a40x40⇒ 1−x+x220=a0x40−a1x39+a2x38−⋯+a40     ......(3)Clearly,           a02−a12+a22+⋯+a402 is the coefficient of x40 in 1+x+x2201−x+x220= Coefficient of x40in⁡1+x2+x420In 1+x2+x420, replace  x2 by y, then the coefficient of y20 in 1+y+y220 is a20. Hence, a02−a12+a22−⋯+a402=a20or  a02−a12+a22−⋯−a192+a202+−a212+⋯+a402=a20or  2a02−a12+a22−⋯−a192+a202=a20or  a02−a12+a22−⋯−a192=a2021−a20 Let, 1+x+x220=∑r=040 arxr  ………(1)Replacing x by 1/x, we get    1+1x+1x220=∑r=040 ar1xror  1+x+x220=∑r=040 arx40−r       ........(2)Since (1) and (2) are same series, coefficient of xr in (1) : coefficient of xr in (2) ⇒ ar=a40−rIn (1), putting x = 1, we get320=a0+a1+a2+⋯+a40=a0+a1+a2+⋯+a19+a20+a21+an+2+⋯+a40=2a0+a1+a2+⋯+a19+a20 ∵ ar=a40−ror  a0+a1+a2+⋯+a19=12320−a20=12910−a20Also,        a0+3a1+5a2+⋯+81a40 =a0+81a40+3a1+79a39+⋯+39a19+43a21+41a20          =82a0+a1+a2+⋯+a19+41a20=41910−a20+41a20=41×320a02−a12+a22−a32+⋯ suggests that we have to multiply the two expansions.Replacing r by - 1/x in (1), we get1−1x+1x220=a0−a1x+a2x2−⋯+a40x40⇒ 1−x+x220=a0x40−a1x39+a2x38−⋯+a40     ......(3)Clearly,           a02−a12+a22+⋯+a402 is the coefficient of x40 in 1+x+x2201−x+x220= Coefficient of x40in⁡1+x2+x420In 1+x2+x420, replace  x2 by y, then the coefficient of y20 in 1+y+y220 is a20. Hence, a02−a12+a22−⋯+a402=a20or  a02−a12+a22−⋯−a192+a202+−a212+⋯+a402=a20or  2a02−a12+a22−⋯−a192+a202=a20or  a02−a12+a22−⋯−a192=a2021−a20