Find the Coefficient of x^{8} in the expansion of ≤ft(b x²+1/b x)¹3}

Correct Answer is : 13C6b¹3}C₆ b13C6b

Correct Answer is : 13C6b¹3}C₆ b13C6b

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UPSC NDA Math Model Paper 8

Question : 78 of 120

Marks: +1, -0

x^{8}

\left(b x^{2}+\frac{1}{b x}\right)^{13}

Solution:

For the given expression

\left(b x^{2}+\frac{1}{b x}\right)^{13}

, n = 13

T_{r+1}=^{13}C_{r}(bx)^{13-r}\left(\frac{1}{bx}\right)^{r}

T_{r+1}=^{13}C_{r} b^{13-r} x^{13-r} \frac{1}{b^{r} x^{r}}

T_{r+1}=^{13}C_{r} b^{13-2r} x^{26-3r}

Now, in order to find out coefficient of

x^{8}, 26-3r

must be 8.

i.e.

26-3r=8

r=6

Hence putting

r=6

in equation (1) we get,

T_{7}=^{13}C_{6} b x^{8}

T_{7}=^{13}C_{6} (bx)^{8}

Therefore of coefficient of

x^{8}

is equal to

^{13}C_{6} b

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