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What is the middle term in the expansion of `(1-(x)/(2))^(8)` ?A. `(35x^(4))/(8)`B. `(17x^(5))/(8)`C. `(35x^(5))/(8)`D. None of these

Answer» Correct Answer - A
Since n = 8 is even number therefore middle term
`=((n)/(2)+1)^(th)"term"=(4+1)=5^(th)"term"`
Hence, `T_(5)=.^(8)C_(4)(1)^(4)(-(x)/(2))^(4)`
`=(8!)/(4!4!)xx(x^(4))/(16)=(8xx7xx6xx5)/(4xx3xx2xx1).(x^(4))/(16)=(70x^(4))/(16)=(35x^(4))/(8)`


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