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Q.

Let r=020arxr=1+x+x210 then a510C1×a4+10C2×a310C3×a2+10C4×a110C5×a0=

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a

0

b

10 C 4  

c

10 C 5  

d

10 C 2    

answer is A.

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Detailed Solution

1+x+ x 2 10 = a 0 + a 1 x+ a 2 x 2 +. (1x) 10 =1 10 C 1 x+ 10 C 2  x 2  10 C 3  x 3 +..  
multiply those two Equation , we get the given  series and is the coefficient   x 5 in 1+ x 3 10  which is zero   

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Let ∑r=020 arxr=1+x+x210 then a5−10C1×a4+10C2×a3−10C3×a2+10C4×a1−10C5×a0=