Question

Find the Coefficient of x⁸ in the expansion of \(\rm \left ( bx^2+\frac{1}{bx} \right )^{13}\)
1. ¹³C₆ b
2. 13C6
3. 13C₈ b
4. 13C8
5. None of these

Answer 1

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

\(\rm T_{r+1}=^{13}\textrm{C}_{r}(bx)^{13-r}\left ( \frac{1}{bx} \right )^r\)

\(\rm T_{r+1}=^{13}\textrm{C}_{r}b^{13-r}(x)^{13-r}\left ( \frac{1}{b^rx^r} \right )\)

\(\rm T_{r+1}=^{13}\textrm{C}_{r}b^{13-2r}(x)^{26-3r}\)

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