Alg10.7

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Find the independent term of x\, in the expansion of (6x^2-\frac{5}{x^2})^6\,

T_{r+1}=6\!C_r\cdot (6x^2)^{6-r}\cdot (\frac{5}{x^2})^r\,

6\!C_r \cdot 6^{6-r}\cdot x^{12-2r}\cdot \frac{5^r}{x^2r}\,

6\!C_r \cdot 6^{6-r}\cdot 5^r \cdot x^{12-2r-2r}\,

To get the constant term to find the value of r so that 12-4r=0,r=3\,

Hence the value independent of x is 4th term T_4=6\!C_3\cdot 6^3 \cdot 5^3\,


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