1
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Question

# If (1+x)10=a0+a1x+a2x2+......a10x10, then value of (a0âˆ’a2+a4âˆ’a6+a8âˆ’a10)2+(a1âˆ’a3+a5âˆ’a7+a9)2 is

A
210
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B
2
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C
220
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D
None of these
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Solution

## The correct option is B None of thesex=i(1+i)10=a0+a1i−a2−a3i+a4......a10 ……….(1)x=−i(1−i)10=a0−a1i−a2+a3i+a4.....a10 ……..(2)(1) + (2)⇒(1+i)10+(1−i)10=2(a0−a2+a4......a10)(√2)10((eiπ/4)10+(e−iπ/4)10)=2(a0+a2+a4....)25=2(a0−a1...…)(a0+a2+a4......a10)2=0(1) - (2)(1+i)10−(1−i)10=(a1−a3+a5...…..an)i(√2)10 (2sin10π4)i=(a1−a3+a5.....a9)i(26)2=(a1−a3+a5.....a9)2=212(a0−a2+a4.....a10)2+(a1−a3+a5...…+a9)2=212.

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