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7.Binomial Theorem
medium
यदि ${\left( {{x^2} + \frac{k}{x}} \right)^5}$ के विस्तार में $x $ का गुणांक $270$ हो, तो $k =$
A
$1$
B
$2$
C
$3$
D
$4$
Solution
${T_{r + 1}} = {}^5{C_r}{({x^2})^{5 – r}}{\left( {\frac{k}{x}} \right)^r}$
$x$ के गुणांक के लिए $10 – 2r – r = 1\,\,\,\, \Rightarrow r = 3$
अत: ${T_{3 + 1}} = {}^5{C_3}{({x^2})^{5 – 3}}{\left( {\frac{k}{x}} \right)^3}$
प्रश्नानुसार, $10\,{k^3} = 270\,\, \Rightarrow \,\,k = 3$.
Standard 11
Mathematics
