UPSEE Solved Model Paper 6

Here we can find the coefficient of x4 in the given expression. We know that Binomial theorem for Positive Integral index If a, b, are any two real numbers and n any natural number then $(a+b)^n$ = $\,,^{n}C_0 a^{n} b^{\circ}$ + $\,,^{n}C_1 a^{n-1}b$ + $\,,^{n}C_2 a^{n-2} b^2$ + ... + $\,,^{n}C_r a^{n-r} b^r$ + ... + $\,,^{n}C_n x^0 b^n$ , where $\,,^{n}C_r$ = $\frac{n!}{(n-r)!r!}$ Here $\,,^nC_0,\,{}^nC_1,\,{}^nC_2$ ... $\,,^{n}C_n$ are called binomial coefficients i.e., $(a+b)^n$ = $\sum^{\underset{r=0}{n}} \,,^{n}C_r a^{n-r} b^r$ Given $(1+x+x^2)^3$ Considering the expansion of $(1 + x + x^2)^3$ in the form $((1 + x) + x^2)^3$ We have $(1 + x)^3 + 3(1 + x)^2 x^2 + 3(1 + x)x^4 + x^6$. (Since, by definition of binomial theorem) Here $x^4$ comes in the second and third term only. Adding its coefficients, 3 + 3 = 6. Therefore, The coefficient of x4 in the expansion of $(1 + x + x^2)^3$ is 6.