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Let m and n be the coefficients of seventh and thirteenth terms respectively in the expansion of ( 13x 13+ 12x 23 )18. Then ( nm ) 13 is :
(4) 94
Tr+1=18Cr ( x1/33 )18-r ( 12x2/3 )r
m (from T7, r=6) =18C6 1312 126
n (from T13, r=12) =18C12 136 1212
18C12=18C6: nm= 312 2636 212= ( 32 )6
( nm ) 13= ( 32 )2= 94
Answer: (4)
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