( a + b ) 0 = 1 ( a + b ) 1 = a + b ( a + b ) 2 = a 2 + 2 a b + b 2 ( a + b ) 3 = ( a + b ) ( a + b ) 2 = ( a + b ) ( a 2 + 2 a b + b 2 ) = a 3 + 2 a 2 b + a b 2 + b a 2 + 2 a b 2 + b 3 = a 3 + 3 a 2 b + 3 a b 2 + b 3 ( a + b ) 4 = ⋯ {\displaystyle {\begin{aligned}(a+b)^{0}&=1\\(a+b)^{1}&=a+b\\(a+b)^{2}&=a^{2}+2ab+b^{2}\\(a+b)^{3}&=(a+b)(a+b)^{2}=(a+b)(a^{2}+2ab+b^{2})\\&=a^{3}+2a^{2}b+ab^{2}+ba^{2}+2ab^{2}+b^{3}\\&=a^{3}+3a^{2}b+3ab^{2}+b^{3}\\(a+b)^{4}&=\cdots \end{aligned}}}