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Question

If (1+x+2x2)20=a0+a1x+a2x2+........+a40x40 then find the value of a0+a2+......+a38

A
219(219+1)
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B
219(2191)
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C
219(2201)
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D
219(220+1)
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Solution

The correct option is C 219(2201)
(1+x+2x2)20=a0+a1x+a2x2+........+a40x40 ....(1)
Put x=1 in (1)
420=a0+a1+a2+........+a40
240=a0+a1+a2+........+a40 ....(2)
Put x=1 in (1)
220=a0a1+a2a3........a39+a40 ....(3)
Adding (2) and (3), we get
240+220=2(a0+a2+a4+......+a38+a40)
220(220+1)=2(a0+a2+a4+......+a38+a40)
219(220+1)=a0+a2+a4+......+a38+a40
a0+a2+a4+......+a38=219(220+1)a40 .....(4)
Since, a40 is the coefficient of x40 in the expansion of (1+x+2x2)20
a40=220
Put this value in (4), we get
a0+a2+a4+......+a38=219(220+1)220
a0+a2+a4+......+a38=219(2201)

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