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Q.

If  ai (i = 0, 1, 2, . . . 16)are real constants such that for every real value of x 1+x+x28=a0+a1x+a2x2+…+a16x16 then a5 is  equal to

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answer is 2.

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Detailed Solution

1+x+x28=∑p,q,r≥0P+q+r=8 8!p!q!r!1pxqx2rFor , a5we require, q + 2r = 5⇒ q=5,r=0,p=3or  q=3,r=1,p=4or q=1,r=2,p=5.Thus, coefficient of x5 is8!5!3!+8!3!1!4!+8!1!2!5!=504
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If  ai (i = 0, 1, 2, . . . 16)are real constants such that for every real value of x 1+x+x28=a0+a1x+a2x2+…+a16x16 then a5 is  equal to