For x= 57, if t_k is the first negative term in the expansion of (1+x )^7/5, then t_1+t_2+ …t_k=

Question

Accepted Answer

Correct answer: C. (c) Given expression is (1+x )^7/5 Apply the binomial expression of (1+x )^n (1+x )^7/5=1+ 75x+ 12 ! 75( 75-1)x^2+ 13 ! ( 75 )( 75-1)( 75-2)x^3 t_4<0 ∴t_1+t_2+t_3+t_4 Put x= 57 in the expansion =1+ 75 ( 57 )+ 12 75 25 ( 57 )^2+ 16 ( 75 ) ( 25 ) ( -35 ) ( 57 )= 10449

$For x = \frac{5}{7}, if t_{k} is the first negative term in the expansion of (1 + x)^{7/5}, then t_{1} + t_{2} + \dots t_{k} =$

Solution

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