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Question

If 3x22x+4(x1)2=A1x+1+A2(x+1)2+A3(x+1)3+A4(x+1)4+A5(x+1)5+A6(x+1)6, then (A1+A3+A5,A2+A4+A6)=

A
(0,0)
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B
(8,12)
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C
(8,12)
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D
(8,12)
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Solution

The correct option is D (8,12)
A1x+1+A2(x+1)2+A3(x+1)3+A4(x+1)4+A5(x+1)5+A6(x+1)6=3x22x+4(x1)6 .... (1)

Now, put x=0 in (1)
A1+A2+A3+A4+A5+A6=4(2)
If A1+A3+A5=8 and A2+A4+A6=12 satisfies the equation
Hence, only option D i.e. (8,12) satisfies the solution.

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