Question

expand (x+3)^5 the binomial theorem

Answer 1

$\underline{\textbf{Given:}}$

$\mathsf{(x+3)^5}$

$\underline{\textbf{To expand:}}$

$\mathsf{(x+3)^5}$

$\underline{\textbf{Solution:}}$

$\underline{\textbf{Binomial theorem:}}$

$\boxed{\mathsf{(a+b)^n=n_{C_0}\,a^n\,b^0+n_{C_1}\,a^{n-1}\,b^1+\;.\;.\;.\;.\;.+n_{C_n}\,a^0\,b^n}}$

$\mathsf{Consider,}$

$\mathsf{(x+3)^5}$

$\textsf{Using binomial theorem,}$

$\mathsf{=5_{C_0}\,x^5\,3^0+5_{C_1}\,x^4\,3^1+5_{C_2}\,x^3\,3^2+5_{C_3}\,x^2\,3^3+5_{C_4}\,x^1\,3^4+5_{C_5}\,x^0\,3^5}$

$\mathsf{=5_{C_0}\,x^5\,3^0+5_{C_1}\,x^4\,3^1+5_{C_2}\,x^3\,3^2+5_{C_2}\,x^2\,3^3+5_{C_1}\,x^1\,3^4+5_{C_5}\,x^0\,3^5}$

$\mathsf{=1\,x^5(1)+5\,x^4(3)+\dfrac{5{\times}4}{1{\times}2}\,x^3(9)+\dfrac{5{\times}4}{1{\times}2}\,x^2(27)+5\,x^1(81)+1\,x^0(243)}$

$\mathsf{=x^5+15\,x^4+90\,x^3+270\,x^2+405x^1+243}$

$\mathsf{=x^5+15\,x^4+90\,x^3+270\,x^2+405\,x+243}$