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Is the co-efficient of $x^{2}$ in the expansion of $(K+bx)^{2}$ is equal to the coefficient of $x^{6}$ in the expansion of $(Kx+b)^{8}$ , then find the value of $K^{6}$

Option 1)
$\frac{15}{28}$

Option 2)
$\frac{15}{14}$

Option 3)
$\frac{1}{28}$

Option 4)
$\frac{5}{28}$

Answers (1)

As we learnt in

General Term in the expansion of (x+a)^n -

$T_{r+1}= ^{n}c_{r}\cdot x^{n-r}\cdot a^{r}$

- wherein

Where $r\geqslant 0 \, and \, r\leqslant n$

$r= 0,1,2,----n$

$\left ( K+bx \right )^{2}=K^{2}+b^{2}x^{2}+(2Kb)x$

Coeffecient of $x^{2}=b^{2}$

$\left ( Kx+b \right )^{8}$

Coeffecient of $x^{6}$ is $^{8}C_{6}\left (Kx \right )^{6}b^{2}=^{8}C_{6}K ^{6}b^{2}x^{6}$

Coeffecient of $x^{6}\, \, is\, \,$ $^{8}C_{6}K ^{6}b^{2}$

Thus here

$\\b^{2}=^{8}C_{6}K ^{6}b^{2}\\*\\* K^{6}=\frac{1}{28}$

Option 1)

$\frac{15}{28}$

Incorrect

Option 2)

$\frac{15}{14}$

Incorrect

Option 3)

$\frac{1}{28}$

Correct

Option 4)

$\frac{5}{28}$

Incorrect

Posted by

Vakul

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