有a.maths 唔識.....
2008-03-20 6:09 am
when (2x - 1/x)^10 is expanded in descending powers of x , find the 3rd term of the expansion
回答 (3)
2008-03-20 6:25 am
✔ 最佳答案
using the general term nCr*(a)^(n-r)*(b)^r in this case a=2x , b=-1/x,n=10
notice that higher the n-r is, higher is the power of x in the term is.
as the expansion in arrange in descending powers of x, take r=2
the ans:70*(2x)^7*(-1/x)^3=-11520x^6
2008-03-20 6:16 am
In general n-1 term will be (2x)^(10-n) * (-1/x)^(n) * 10Cn
Hence the 3rd term = (2x)^8 * (-1/x)^2* 10C2 = ( 256x^6)*45= 11520x^6
Hence the 3rd term = (2x)^8 * (-1/x)^2* 10C2 = ( 256x^6)*45= 11520x^6
參考: me
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