If for some positive integer n, the coefficients of three consecutive terms in the binomial expansion of (1 + x)n + 5 are in the ratio 5 : 10 : 14, then theIargest coefficient in this expansion is

If for some positive integer n, the coefficients of three consecutive terms in the binomial expansion of (1 + x)n + 5 are in the ratio 5 : 10 : 14, then the
Iargest coefficient in this expansion is

  1. A

    330

  2. B

    462

  3. C

    792

  4. D

    252

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    Solution:

     Let the three consecutive terms in the binomial expansion of (1+x)n+5 are

     n+5Cr1,n+5Cr and n+5Cr+1 

    Now, according to the given information  n+5Cr1:n+5Cr:n+5Cr+1=5:10:14

     (n+5)!(r1)!(nr+6)!:(n+5)!r!(nr+5)!:(n+5)!(r+1)!(nr+4)!=5:10:14 

     

     1(nr+6)(nr+5):1r(nr+5):1(r+1)r=5:10:14  So,  rnr+6=510  2r=nr+6n+6=3r  and  r+1nr+5=57 7r+7=5n5r+25 5n+18=12r

    From Eqs. (i) and (ii), we have n =6.
    So, the largest coefficient in the expansion is same as 

    the greatest binomial coefficient 

    =11C5 or 11C6=11!5!6!=11×10×9×8×75×4×3×2=462

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