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If coefficient of x3 and x4 in the expansion of 1+ax+bx2(12x)18 in powers of x are both zeros, then (a,b) is equal to 

a
14,2513
b
14,2723
c
16,2723
d
16,2513

detailed solution

Correct option is C

S=1+ax+bx2(1−2x)18=1+ax+bx21+18C1(−2x)+ 18C2(−2x)2+18C3(−2x)3+18C4(−2x)4+…Coefficient of x3 in the expansion of S is  18C3(−2)3+a 18C2(−2)2+18C1(−2)b=0Divide by  18C1(−2) to obtain5443−17a+b=0               (1)Similarly, coefficient of x4 is 18C4(−2)4+a 18C3(−2)3+18C2(−2)2b=0Divide by  18C2(−2)2 to obtain80−323a+b=0                 (2)Subtract (2) from (1) to obtain3043−193a=0⇒a=16From b=17×16−5443=2723

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