Can you please solve this question along with telling general approach for such pattern of questions?
Q.5. If 1 + x + x 2 + x 3 x 5 = a k k = 0 15           x k   t h e n   a 2 k k = 0 7               is equal to
(a) 128
(b) 256
(c) 512
(d) 1024

Dear Student,
Please find below the solution to the asked query:

Each question requires different apporach. But expansions withmore than 2 terms are rarely asked even in exams like JEEAdvanced.Now1+x+x2+x35=k=015 akxk1+x+x2+x35=a0+a1x+a2x2+a3x3+....+a14x14+a15x15...iPut x=11+1+1+15=a0+a1+a2+a3+....+a14+a1545=a0+a1+a2+a3+....+a14+a15.....iiPut x=-1 in i1-1+1-15=a0-a1+a2-a3+....+a14-a150=a0-a1+a2-a3+....+a14-a15....iiiii+iiia0+a1+a2+a3+....+a14+a15+a0+a1+a2+a3+....+a14+a15=452a0+a2+....+a14=1024a0+a2+....+a14=512k=07 a2k=512 Answer

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It is sigma in the question with akxk part
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