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)
V Vakul

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

Exams
Articles
Questions