If (1+x+x2)(1−x1!+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 Aa2=12 (1+x+x2)(1−x1!+x22!−x33!+)=a0+a1x+a2x2+a3x3+a4x4+...a1=1−1=0a2=12−1+1=12a3=−16+12−1=−23a4=124−16+12=38 Hence, option 'B' is correct.