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Question

If (1+x+x2)(1x1!+x22!x33!+)=a0+a1x+a2x2+a3x3+a4x4+... then,

A
a1=1
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B
a2=12
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C
a3=13
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D
a4=12
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Solution

The correct option is A a2=12
(1+x+x2)(1x1!+x22!x33!+)=a0+a1x+a2x2+a3x3+a4x4+...a1=11=0a2=121+1=12a3=16+121=23a4=12416+12=38
Hence, option 'B' is correct.

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