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7.Binomial Theorem
normal
$(x^2 - x + 1)^{10} (x^2 + 1 )^{15}$ ના વિસ્તરણમાં $x^3$ નો સહગુણક મેળવો
A
$-360$
B
$-240$
C
$-180$
D
$60$
Solution
$\left(x^{2}-x+1\right)^{10} \cdot\left(x^{2}+\right)^{15}$
$=\left(1+^{10} \mathrm{C}_{1}\left(\mathrm{x}^{2}-\mathrm{x}\right)+^{10} \mathrm{C}_{2}\left(\mathrm{x}^{2}-\mathrm{x}\right)^{2}+^{10} \mathrm{C}_{3}\left(\mathrm{x}^{2}-\mathrm{x}\right)^{3}\right)$
$\left(^{15} \mathrm{C}_{0}+^{15} \mathrm{C}_{1}, \mathrm{x}^{2}\right)$
$=-2 \cdot^{10} \mathrm{C}_{2} \cdot^{15} \mathrm{C}_{0}-^{10} \mathrm{C}_{1} \cdot^{15} \mathrm{C}_{1}-^{10} \mathrm{C}_{3}=-360$
Standard 11
Mathematics