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7.Binomial Theorem
easy
વિસ્તરણનું વ્યાપક પદ લખો : $\left(x^{2}-y x\right)^{12}, x \neq 0$
Option A
Option B
Option C
Option D
Solution
It is known that the general term ${T_{r + 1}}{\rm{ \{ }}$ which is the ${(r + 1)^{{\rm{th }}}}$ term $\} $ in the binomial expansion of $(a+b)^{n}$ is given by ${T_{r + 1}} = {\,^n}{C_r}{a^{n – r}}{b^r}$
Thus, the general term in the expansion of $\left(x^{2}-y x\right)^{12}$ is
${T_{r + 1}} = {\,^{12}}{C_r}{\left( {{x^2}} \right)^{12 – r}}{( – yx)^r} = {( – 1)^r}{\,^{12}}{C_r} \cdot {x^{24 – 2r}}{y^r} = {( – 1)^r}{\,^{12}}{C_r} \cdot {x^{24 – r}} \cdot {y^r}$
Standard 11
Mathematics
