find the coefficient of x7 in ( ax2 + 1/bx)11 and the coefficient of x-7 in (ax - 1/bx2)11 . if these coefficients are equal then find the relation between a and b

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Please find below the solution to the asked query:

We know that r+1th term in expansion of x+an is given by:Tr+1=nCr.xn-r.arWe have ax2+1bx11Let x7 occur in r+1th term, henceTr+1=11Crax211-r1bxr=11Cra11-rbrx22-2rxr=11Cra11-rbrx22-2r-r=11Cra11-rbrx22-3rAs this term contains x722-3r=73r=15r=5Hence coefficient of x7 in ax2+1bx11Coefficient of  x7 =11C5a11-5b5Coefficient of  x7=11C5a6b5ax-1bx211Let x-7 occurs in r+1th term, henceTr+1=11Cr ax11-r-1bx2r=-1r 11Cr a11-rbrx11-rx2r=-1r 11Cr a11-rbrx11-r-2r=-1r 11Cr a11-rbrx11-3rAs this term contains x-7, hence,11-3r=-73=18r=6Hence coefficient of x-7 in ax-1bx211=-16 11C6 a11-6b6C2As nCr=nCn-rcoefficient of x-7 in ax-1bx211=11C11-6 a5b6coefficient of x-7 in ax-1bx211=11C5 a5b6Given that:coefficient of x7 in ax2+1bx11 = coefficient of x-7 in ax-1bx21111C5a6b5=11C5 a5b6a6b5=a5b6a6a5=b5b6a=1b ab=1 

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