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Find the Coefficient of x8 in the expansion of \(\rm \left ( bx^2+\frac{1}{bx} \right )^{13}\)1. 13C6 b2. 13C63. 13C8 b4. 13C8 |
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Answer» Correct Answer - Option 1 : 13C6 b Concept: The general term of the expansion of (x + a)n is usually denoted by Tr+1 = nCr xn-r ar Calculation: For the given expression \(\rm \left ( bx^2+\frac{1}{bx} \right )^{13}\) , n = 13 \(\rm T_{r+1}=^{13}\textrm{C}_{r}(bx^2)^{13-r}\left ( \frac{1}{bx} \right )^r\) \(\rm T_{r+1}=^{13}\textrm{C}_{r}b^{13-r}(x^2)^{13-r}\left ( \frac{1}{b^rx^r} \right )\) \(\rm T_{r+1}=^{13}\textrm{C}_{r}b^{13-2r}(x)^{26-3r}\) Now, in order to find out coefficient of x8, 26 - 3r must be 8. i.e. 26 - 3r = 8 r = 6 Hence putting r = 6 in equation (1) we get, \(\rm T_{6+1}=^{13}\textrm{C}_{6}b^{13-12}(x)^{26-18}\) \(\rm T_{6+1}=^{13}\textrm{C}_{6}bx^{8}\) Therefore of coefficient of x8 is equal to 13C6 b |
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