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19th question

19th question •f let value N. then the value - 2x l. (f a. lies tu•tween the tind the smallest positive integral of in (l + + + 414)'. resB•ctively •men the • is divided by 49. then the or the numtx•r N — 3t@' is. 7. 8. the greatest Integer prove that then show that b. Prove that where an even integer. Let n tx• a vx»sitive integer and (l + x + x2r Show thata: —a: + a: + -a. coeßicien\ of e 2S9 where n is an ese

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Please find below the solution to the asked query:

We have1+x+x2n=a0+a1x+a2x2+...anxn,,,.+a2nx2n...iReplace x by  -1x⇒1-1x+1x2n=a0-a1x+a2x2+...+a2nx2n...iiNow a02-a12+a22-.....+a2n2 will be generated when we multiplyand equate coeffcient of x0 term independent of x becausea0×a0=a02, -a1x×a1x=-a12 and so on, hence multiply i and ii and equate coefficient of x0⇒Coefficient of x0 in 1+x+x2n×1-1x+1x2n=a02-a12+a22-.....+a2n2...iii Coefficient of x0 in 1+x+x2n×1-1x+1x2n=Coefficient of x0 in 1+x+x2n×x2-x+1x2n=Coefficient of x0 in 1+x2+xn×1+x2+xnx2n=Coefficient of x0 in  1+x22-x2nx2n=Coefficient of x0 in  1+x4+2x2-x2nx2n=Coefficient of x0 in  1+x2+x4nx2n=Coefficient of x2n in 1+x2+x4nAs denominator has also  x2n, hence x2n will cancel out =Coefficient of tn in 1+t+t2n Put x2=t=an Using i as coefficient does not change on changing variable Hence by iiia02-a12+a22-.....+a2n2=an Proved

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